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Lesson 1: Reciprocal, Quotient, and Pythagorean Identities Specific Outcome 6: Prove trigonometric identities, using: reciprocal identities, quotient identities, Pythagorean identities What is the difference between an equation and an identity? Talk through this idea with the students and see if they can come up with the following ideas. 2๐ฅ 2 + 3 = 11 is an equation. It is only true for certain values of the variable x. The solutions to this equation are -2 and 2 which can be verified by substituting these values into the equation. (๐ฅ + 1)2 = ๐ฅ 2 + 2๐ฅ + 1 is an identity. It is true for all values of the variable x. Comparing Two Trigonometric Expressions 1. Graph the curves y = sin x and y = cos x tan x over the domain โ360° โค ๐ฅ < 360°. Graph the curves on separate grids using the same range and scale. What do you notice? They are the same graph. 2. Use the table function on your calculator to compare the two graphs. Describe your findings. Same values, except for the non-permissible values. 3. Use your knowledge of tan x to simplify the expression cos x tan x ๐ ๐๐๐ฅ cos ๐ฅ cos ๐ฅ The cos x cancel off to give sin x. 4. What are the non-permissible values of x in the equation sin x = cos x tan x? Remember that we cannot divide by 0? So use that info to find non-permissible sin ๐ฅ values. Given tan ๐ฅ = cos ๐ฅ, the non-permissible values will occur when cos x = 0. That occurs at 90ห and again at 270ห. (denominator cannot = 0) The equations y = sin x and y = cos x tan x are examples of trigonometric identities. Trigonometric identities can be verified both numerically and graphically. Reciprocal Identities: 1 1 csc x ๏ฝ sec x ๏ฝ sin x cos x Quotient Identities: sin x tan x ๏ฝ cos x cot x ๏ฝ cot x ๏ฝ 1 tan x cos x sin x Verifying Identities for a Particular Case: (plugging in values) When verifying an identity we must treat the left side (LS) and the right side (RS) separately and work until both sides represent the same value. This technique does not prove that an identity is true for all values of the variable โ only for the value of the variable being verified. sin x . cos x a. Determine the non-permissible values, in degrees, for the equation. Ex. Verify that tan x ๏ฝ Occurs when cos x = 0 (at 90ห and 270ห), and can be represented as ๐ โ 90° + 180°๐ (general solution for NPV) b. Verify that ฮธ= 60ห. L.S. R.S. sin 60๏ฐ cos 60๏ฐ tan 60๏ฐ 3 ๏ฝ 3 (using unit circle or triangles) ๏ฝ 1 2 2 3 2 ๏ด 2 1 ๏ฝ 3 ๏ฝ LS = RS c. Verify graphically that this is an identity. Answer: When you graph each side, they are the same graph. 2 Ex. a. Determine the non-permissible values, in radians, of the variable in the sec ๐ฅ expression tan ๐ฅ The functions sec x and tan x both have non-permissible values in their domains: tan ๐ฅ = sin ๐ฅ ๐ , ๐กโ๐๐๐๐๐๐๐ cos ๐ฅ โ 0, (๐ฅ โ + ๐๐) cos ๐ฅ 2 1 sec ๐ฅ = = ๐ ๐๐๐ ๐๐ ๐๐๐๐ฃ๐. cos ๐ฅ The value of tan x also cannot equal 0. Tan ฮธ =0 at ฯ ๐ Therefore, combine the restriction is ๐ฅ โ 2 ๐. A restriction will occur every 90ห. b. Simplify the expression (substitute known identities into the equation) 1 cos ๐ฅ sin ๐ฅ cos ๐ฅ 1 cos ๐ฅ cos ๐ฅ 1 × = = = ๐๐ ๐๐ฅ cos ๐ฅ sin ๐ฅ cos ๐ฅ sin ๐ฅ sin ๐ฅ Pythagorean Identity Recall that point P on the terminal arm of an angle ฮธ in standard position has coordinates (cos ฮธ, sin ฮธ). Consider a right triangle with a hypotenuse of 1 and legs of cos ฮธ and sin ฮธ. y P(cosฮธ, sinฮธ) x The hypotenuse is 1 because it is the radius of the unit circle. Apply the Pythagorean Theorem in the right triangle to establish the Pythagorean Identity: ๐ฅ2 + ๐ฆ2 = 1 ๐๐๐ 2 ๐ + ๐ ๐๐2 ๐ = 1 Pythagorean Identities: sin 2 x ๏ซ cos 2 x ๏ฝ 1 1 ๏ซ tan 2 x ๏ฝ sec2 x 1 ๏ซ cot 2 x ๏ฝ csc2 x 3 These identities can be written in several ways: sin 2 x ๏ฝ 1 ๏ญ cos 2 x tan 2 x ๏ฝ sec2 x ๏ญ 1 cos 2 x ๏ฝ 1 ๏ญ sin 2 x cot 2 x ๏ฝ csc2 x ๏ญ 1 Ex. Verify the identity 1 + ๐๐๐ก 2 ๐ฅ = ๐๐ ๐ 2 for the given values: a. ๐ฅ = ๐ 6 ๐ 2 ๐ 2 L.S. = 1 + (๐๐๐ก 6 ) R.S. = (๐๐ ๐ 6 ) 2 = 22 =4 = 1 + (โ3) =1+3 =4 4๐ b. 3 240ห(QIII) will have a ref angle of 60ห, therefore the Tan will be positive and sin will be negative. tan 60 = โ3, sin 60 = L.S. = 1 + (๐๐๐ก 4๐ 2 ) 3 2 โ3 2 R.S. = (๐๐ ๐ 1 =1+( ) โ3 1 =1+ 2 (โ3) 1 =1+ 3 4 = 3 =( โ2 โ3 4 =3 4๐ 2 ) 3 2 ) b. Use quotient identities to verify the Pythagorean identity 1 + ๐๐๐ก 2 ๐ฅ = ๐๐ ๐ 2 x ๐๐๐ก 2 ๐ฅ + 1 = ๐๐ ๐ 2 ๐ฅ ๐๐๐ 2 ๐ฅ 1 + 1 = ๐ ๐๐2 ๐ฅ ๐ ๐๐2 ๐ฅ 2 2 ๐๐๐ ๐ฅ ๐ ๐๐ ๐ฅ 1 + = 2 2 ๐ ๐๐ ๐ฅ ๐ ๐๐ ๐ฅ ๐ ๐๐2 ๐ฅ ๐๐๐ 2 ๐ฅ + ๐ ๐๐2 ๐ฅ 1 = 2 ๐ ๐๐ ๐ฅ ๐ ๐๐2 ๐ฅ 1 1 = 2 ๐ ๐๐ ๐ฅ ๐ ๐๐2 ๐ฅ ๐๐ ๐ 2 ๐ฅ = ๐๐ ๐ 2 ๐ฅ Assignment: pg. 296-297 #1, 3, 4, 6, 9 โ 11, 13 โ17, C2 4 Lesson 2: Sum, Difference and Double-Angle Identities Specific Outcome 6: Prove trigonometric identities, using: โข โข โข โข โข reciprocal identities quotient identities Pythagorean identities sum or difference identities (restricted to sine, cosine and tangent) double-angle identities (restricted to sine, cosine and tangent). Ex. Use exact values to verify which of the following statements are true: a. sin(60 + 30)º= sin60º + sin30º b. sin ๏จ 60 ๏ซ 30๏ฉ ๏ฐ ๏ฝ sin 60๏ฐ cos30๏ฐ ๏ซ sin 30๏ฐ cos 60๏ฐ Have the students go through both examples using exact value and see which one is true. Once theyโve figured it out, go through the different sum and difference identities. Sum and Difference Identities: sin( A ๏ซ B) ๏ฝ sin A cos B ๏ซ cos A sin B sin( A ๏ญ B) ๏ฝ sin A cos B ๏ญ cos A sin B cos( A ๏ซ B) ๏ฝ cos A cos B ๏ญ sin A sin B cos( A ๏ญ B) ๏ฝ cos A cos B ๏ซ sin A sin B tan A ๏ซ tan B tan( A ๏ซ B) ๏ฝ 1 ๏ญ tan A tan B tan A ๏ญ tan B tan( A ๏ญ B) ๏ฝ 1 ๏ซ tan A tan B Thankfully, these are on the formula sheet so they do not need to be memorized. However, we can use them in a number of problems. 1. The first type is simplifying expressions into single trigonometric functions: Ex. Write the expression cos 88° cos 35° + sin 88° sin 35° as a single trigonometric function. Look on the formula sheet. It has the same form as the right side of cos(A โ B): Therefore we can substitute 88 for A and 35 for B. cos (88-35) = cos53ห 2. The second type is finding exact values of angles that are not on our special triangle. 5 Ex. Find the exact value of sin15๏ฐ The first step is to look at our special triangle angles: 30, 45 and 60. Then see which combinations break the angle into exact values sin15๏ฐ ๏ฝ sin ๏จ 45 ๏ญ 30๏ฉ ๏ฐ ๏ฝ sin 45๏ฐ cos30๏ฐ ๏ญ sin 30๏ฐ cos 45๏ฐ 2 3 1 2 ๏ท ๏ญ ๏ท 2 2 2 2 6 2 ๏ฝ ๏ญ 4 4 ๏ฝ 6๏ญ 2 4 Ex. Find the exact value of ๐ก๐๐ 11๐ 12 When faced with radians, it may be easier to switch to degrees (165ห). Otherwise you will have to find common denominators. First step is to try the special triangle angles or use unit circle. If those donโt work, then we also need to look at the angles that are in other quadrants. If you are looking at degrees, we can use 135 and 30. Rewrite as a addition of two special angles (135 + 30) tan ๐ด + tan ๐ต 1 โ tan ๐ด tan ๐ต tan 135 + tan 30 tan(135 + 30) = 1 โ tan 135 tan 30 tan(๐ด + ๐ต) = From special triangles tan 30 = โ3 3 For 135, it has a reference angle of 45 in QII (tan is negative). Tan 45 = 1 1 โ3 tan(135 + 30) = 1 1 โ (โ1 ( )) โ3 1 โ1 + โ3 tan(135 + 30) = 1 1+ โ3 โ1 + 6 1 โ3 + tan(135 + 30) = โ3 โ3 1 โ3 + โ3 โ3 (common denominators) Simplify by combining fractions: โโ3 + 1 โ3 + 1 tan(135 + 30) = ÷ โ3 โ3 โโ3 + 1 โ3 tan(135 + 30) = × โ3 โ3 + 1 โโ3 + 1 tan(135 + 30) = โ3 + 1 โ There are also double angle identities. Have the students try to find the identities for sin2A and cos2A by using the sum and difference identities. Once we have gone through this, give the identities to the students. Double Angle Identities: sin 2 A ๏ฝ 2sin A cos A cos 2 A ๏ฝ cos 2 A ๏ญ sin 2 A 2 tan A tan 2 A ๏ฝ 1 ๏ญ tan 2 A Note: cos2A has two other forms. Go through these with the students as well. Ex. Determine an identity for cos 2A that contains only the sine ratio. cos 2๐ด = ๐๐๐ 2 ๐ด โ ๐ ๐๐2 ๐ด cos 2๐ด = (1 โ ๐ ๐๐2 ๐ด) โ ๐ ๐๐2 ๐ด (Pythagorean identity) cos 2๐ด = 1 โ 2๐ ๐๐2 ๐ด Ex. Determine an identity for cos 2A that contains only the cosine ratio. cos 2๐ด = ๐๐๐ 2 ๐ด โ ๐ ๐๐2 ๐ด cos 2๐ด = ๐๐๐ 2 ๐ด โ (1 โ ๐๐๐ 2 ๐ด) cos 2๐ด = 2๐๐๐ 2 ๐ด โ 1 These identities are on the formula sheet 7 Ex. Express each in terms of a single trigonometric function: a. 2 sin 4x cos 4x This follows (2 sinA cosA) = sin 2(4x)=sin 8x 1 1 A ๏ญ sin 2 A 2 2 ๏ฆ1 ๏ถ cos 2 ๏ง A ๏ท ๏จ2 ๏ธ ๏ฝ cos A b. cos 2 5 5 c. sin x cos x 2 2 1๏ฆ 5 5 ๏ถ ๏ง 2sin x cos x ๏ท 2๏จ 2 2 ๏ธ 1 ๏ฆ5 ๏ถ ๏ฝ sin 2 ๏ง x ๏ท 2 ๏จ2 ๏ธ 1 ๏ฝ sin 5 x 2 (the 1/2 is added to the outside of the brackets to balance off the 2 added to the inside, the 2 is added to match the identity sin2A=2 sinAcosA) One of the most common questions is to simplify expressions using identities. sin 2๐ฅ Ex. Simplify the expression cos 2๐ฅ+1 to one of the three primary trigonometric ratios 2 sin ๐ด cos ๐ด (2๐๐๐ 2 ๐ด โ 1) + 1 2 sin ๐ด cos ๐ด = 2๐๐๐ 2 ๐ด sin ๐ฅ = = tan ๐ฅ cos ๐ฅ = Assignment: pg. 306-308 #1ae, 2cd, 4bc, 5abd, 7, 8acdf, 10, 11, 15, 16b, 17, 19b, 20ad, 24 8 Lesson 3: Proving Identities Specific Outcome 6: Prove trigonometric identities, using: โข โข โข โข โข reciprocal identities quotient identities Pythagorean identities sum or difference identities (restricted to sine, cosine and tangent) double-angle identities (restricted to sine, cosine and tangent). To prove that an identity is true for all permissible values, it is necessary to express both sides of the identity in equivalent forms. One or both sides of the identity must be algebraically manipulated into an equivalent form to match the other side. You cannot perform operations across the equal sign when proving a potential identity. The equal sign is a border! Simplify the expressions on each side of the identity independently. Hints in Proving an Identity: 1. Begin with the more complex side 2. If possible, use known identities given on the formula sheet, eg. try to use the Pythagorean identities when squares of trigonometric functions are involved. 3. If necessary change all trigonometric ratios to sines and/or cosines, eg. sin x 1 replace tan x by , or sec x by . cos x cos x 4. If there is a sum or difference of fractions, write as a single fraction 5. Look for factoring as a step in trying to prove an identity. 6. Occasionally, you may need to multiply the numerator or denominator of a fraction by its conjugate. It is usually easier to make a complicated expression simpler than it is to make a simple expression more complicated. In this section we will be doing verifying again as well as proving. Make sure that you can do both. Verifying means we simply plug in an angle and prove that left side = right side. Proving means that we need to do it algebraically. 9 Ex. Consider the statement 1 ๏ญ cos x ๏ฝ sin x tan x cos x a. Verify the statement is true for x ๏ฝ L.S. 1 cos ๏ฐ ๏ญ cos ๏ฐ 3 R.S. ๏ฐ 3 sin 3 1 1 ๏ฝ ๏ญ 1 2 2 1 ๏ฝ 2๏ญ 2 3 ๏ฝ 2 ๏ฐ tan ๏ฐ 3 3 3 ๏ฝ ๏ท 3 2 3 ๏ฝ 2 b. Prove the statement algebraically: L.S. 1 ๏ญ cos x cos x 1 ๏ญ cos 2 x ๏ฝ cos x sin 2 x ๏ฝ cos x R.S. sin x tan x sin x ๏ฝ sin x ๏ท cos x 2 sin x ๏ฝ cos x c. State the non-permissible values for x. The only non-permissible values for either side is cos x โ 0. ๐ฅโ ๐ + ๐๐ 2 2 tan ๐ฅ Ex. Consider the identity 1+๐ก๐๐2 ๐ฅ = sin 2๐ฅ a. Describe how to use a graphing calculator to verify the identity. 2 tan ๐ฅ If you graph ๐ฆ1 = 1+๐ก๐๐2 ๐ฅ and graph ๐ฆ2 = sin 2๐ฅ, The graphs should be identical 10 b. Prove the identity L.S. sin ๐ฅ 2 cos ๐ฅ = ๐ ๐๐ 2 ๐ฅ 2 sin ๐ฅ = cos ๐ฅ 1 ๐๐๐ 2 ๐ฅ R.S. = 2 sin ๐ฅ cos ๐ฅ 2 sin ๐ฅ ๐๐๐ 2 ๐ฅ × cos ๐ฅ 1 = 2 sin ๐ฅ cos ๐ฅ = 1 Ex. Prove that 1โsin ๐ด = 1+sin ๐ด ๐๐๐ 2 ๐ด is an identity for all permissible values of x. (this is an example of multiplying by the conjugate. Also show them how to do the identity substituting in for the cos denominator and factoring difference of squares.) ๐ฟ๐ = 2 cos ๐ฅโ1 1 1 + sin ๐ด × 1 โ sin ๐ด 1 + sin ๐ด 1 + sin ๐ด = 1 โ ๐ ๐๐2 ๐ด 1 + sin ๐ด = ๐๐๐ 2 ๐ด 1 Ex. Prove that 2๐๐๐ 2 ๐ฅโ7 cos ๐ฅ+3 = cos ๐ฅโ3 is an identity. For what values of x is this identity undefined? 2 cos ๐ฅ โ 1 โ 6 cos ๐ฅ โ cos ๐ฅ + 3 2 cos ๐ฅ โ 1 = 2 cos ๐ฅ(cos ๐ฅ โ 3) โ 1(cos ๐ฅ โ 3) 2 cos ๐ฅ โ 1 = (cos ๐ฅ โ 3)(2 cos ๐ฅ โ 1) 1 = cos ๐ฅ โ 3 ๐ฟ๐ = 2๐๐๐ 2 ๐ฅ 11 For non-permissible values, we need to find the equations cos x โ 3 = 0 or 2cos x โ 1 = 0 are undefined. 1 2 Quad. I and IV (positive) Ref angle is 60ห QI: 60ห QIV: 360ห-60ห = 300ห Undefined for x = 60ห+360หn, nฮตI x = 300ห+360หn, nฮตI cos ๐ฅ = 3 No solution. cos ๐ฅ = I Assignment: pg. 314-315 #1ad, 2ac, 3acd, 5, 7b, 10ac, 11bc, 12, 14, 18, C1, C2 12 Lesson 4: Solving Trigonometric Equations Using Identities Specific Outcome 5: Solve, algebraically and graphically, first and second degree trigonometric equations with the domain expressed in degrees and radians. For our last section of trig, we are going to combine identities and solving equations. Investigation 1. Graph the function ๐ฆ = sin 2๐ฅ โ sin ๐ฅ over the domain โ720° < ๐ฅ โค 720°. Describe the graph. 2. From the graph, determine an expression for the zeros of the function over the domain of all real numbers. (donโt do allโฆjust find a couple for practice) -660ห, -540ห, -420ห, -360ห, -300ห,-180ห, -60ห, 0ห, 60ห, 180ห, 300ห, 360ห, 420ห, 540ห, 660ห, 720ห 4 solutions for each cycle (look from max to max or min to min) 3. Algebraically solve sin 2๐ฅ โ ๐ ๐๐๐ฅ = 0 over the domain. Compare your answers to the question above. sin 2๐ฅ โ ๐ ๐๐๐ฅ = 0 = 2 sin ๐ฅ cos ๐ฅ โ sin ๐ฅ = sin ๐ฅ (2 cos ๐ฅ โ 1) sin ๐ฅ = 0 At 0ห, 180ห, 360ห, โฆ. ๐ฅ = 0° + 180°๐ 1 2 Ref angle is 60ห(answers in QI and QIV) = 60ห, 300ห, 420ห, 660หโฆ ๐ฅ = 60° + 360°๐ ๐ฅ = 300° + 360°๐ cos ๐ฅ = 4. Which method, graphic or algebraic, do you prefer to solve the above equation? To solve some trigonometric equations, you need to make substitutions using the trigonometric identities that you have studied in this chapter. This often involves ensuring that the equation is expressed in terms of one trigonometric function. 13 Ex. Solve the following equation where 0 ๏ฃ x ๏ฃ 2๏ฐ . (easier to use graphing calculatorโฆknow both) 2 cos 2 x ๏ญ 3sin x ๏ฝ 0 2(1 ๏ญ sin 2 x) ๏ญ 3sin x ๏ฝ 0 ๏ฎ Pythagorean identity 2 ๏ญ 2sin 2 x ๏ญ 3sin x ๏ฝ 0 ๏ฎ Expand brackets 0 ๏ฝ 2sin 2 x ๏ซ 3sin x ๏ญ 2 ๏ฎ Need to factor ๏จ 2sin x ๏ญ1๏ฉ๏จsin x ๏ซ 2๏ฉ sin x ๏ฝ 1 2 sin x ๏ฝ ๏ญ2 Positive in Quad I and II ref ๏ ๏ฝ sin ๏ญ1 (0.5) ๏ฝ 30๏ฐ ๏ฎ x๏ฝ ๏ฐ ๏ฆ , ๏ง๏ฐ 6 ๏จ No solution ๏ฐ 6 ๏ฐ 5๏ฐ ๏ฐ๏ถ ๏ญ ๏ท or x ๏ฝ , 6 6 6๏ธ But, of course, you can use the graphing calculator. Ex. Solve for x as an exact value where 0 โค ๐ฅ โค 2๐ 3 โ 3 sin ๐ฅ โ 2๐๐๐ 2 ๐ฅ 3 โ 3 sin ๐ฅ โ 2(1 โ ๐ ๐๐2 ๐ฅ) = 0 3 โ 3 sin ๐ฅ โ 2 + 2๐ ๐๐2 ๐ฅ = 0 2๐ ๐๐2 ๐ฅ โ 3 sin ๐ฅ + 1 = 0 2๐ ๐๐2 ๐ฅ โ 2 sin ๐ฅ โ sin ๐ฅ + 1 = 0 2 sin ๐ฅ (sin ๐ฅ โ 1) โ 1(sin ๐ฅ โ 1) = 0 (sin ๐ฅ โ 1)(2 sin ๐ฅ โ 1) = 0 1 2 ๐ Quad I, II, ref angle is 6 sin ๐ฅ = 1 ๐ Quad I, II, ref angle is 2 ๐ ๐ฅ= 2 sin ๐ฅ = ๐ฅ= 14 ๐ 5๐ , 6 6 1 Ex. Solve the equation ๐ ๐๐2 ๐ฅ = 2 tan ๐ฅ cos ๐ฅ algebraically over the domain 0° โค ๐ฅ < 360°. 1 sin ๐ฅ ๐ ๐๐2 ๐ฅ = × × cos ๐ฅ 2 cos ๐ฅ 2๐ ๐๐2 ๐ฅ = sin ๐ฅ 2๐ ๐๐2 ๐ฅ โ sin ๐ฅ = 0 sin ๐ฅ (2 sin ๐ฅ โ 1) 1 2 Ref angle is 30ห, pos in QI and II ๐ฅ = 30°, 150° sin ๐ฅ = 0 sin ๐ฅ = ๐ฅ = 180° Check for any non-permissible values: occur when cos x = 0, at 90ห, 270ห The non-permissible values are not part of our solution set, therefore our solutions are x = 30ห, 150ห, and 180ห b. Verify your answer graphically. Ex. Algebraically solve cos 2๐ฅ = cos ๐ฅ. Give general solutions expressed in radians. ๐๐๐ 2 ๐ด โ ๐ ๐๐2 ๐ด = cos ๐ฅ ๐๐๐ 2 ๐ด โ (1 โ ๐๐๐ 2 ๐ด) = cos ๐ฅ 2๐๐๐ 2 ๐ด โ 1 = cos ๐ฅ 2๐๐๐ 2 ๐ด โ cos ๐ฅ โ 1 = 0 2 2๐๐๐ ๐ด โ 2 cos ๐ด + cos ๐ด โ 1 = 0 (2 cos ๐ด + 1)(cos ๐ด โ 1) = 0 1 2 ๐ Ref angle is 3 , neg in QII andQIII ๐ 2๐ ๐ฅ =๐โ = 3 3 ๐ 4๐ ๐ฅ =๐+ = 3 3 ๐ถ๐๐ ๐ด = 1 Occurs at 0 cos ๐ด = โ General Solutions: 2๐ 3 + 2๐๐, 4๐ 3 + 2๐๐, 0 + 2๐๐ 15 Ex. Algebraically solve 3 cos ๐ฅ + 2 = 5 sec ๐ฅ. Give general solutions expressed in radians. 3 cos ๐ฅ + 2 = 5 sec ๐ฅ 1 3 cos ๐ฅ + 2 = 5 × cos ๐ฅ 5 3 cos ๐ฅ + 2 = cos ๐ฅ cos ๐ฅ (3 cos ๐ฅ + 2) = 5 3๐๐๐ 2 ๐ฅ + 2 cos ๐ฅ โ 5 = 0 3๐๐๐ 2 ๐ฅ โ 3 cos ๐ฅ + 5 cos ๐ฅ โ 5 = 0 (cos ๐ฅ โ 1)(3 cos ๐ฅ + 5) = 0 5 3 No solution. cos ๐ฅ = 1 Occurs at 0 cos ๐ฅ = โ Only non-permissible value occurs when cosx = 0, (90ห, 270ห) The non-permissible values do not change our original solution of 0 +2ฯn Unless the domain is restricted, give general solutions. Check that the solutions for an equation do not include non-permissible values from the original equation. Assignment: pg. 320-321 #1ad, 2bc, 3bc, 5, 6, 8, 10 โ 12, 15, 18, C2 16