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8.1 Simple Trig Equations οΌThere are often multiple (infinite) solutions to trigonometric equations. For example take the equation sin(x)=.5. Find the solutions. Understand that this graph repeats itself every interval of 2pi, so there will also be π π π solutions at + ππ and then at + ππ and then at + ππ . π π So we would re-write these infinite answers as π π π + πππ where k is an integer. β’ With the sine and cosine, because the period is 2pi, or the distance that it takes to repeat itself is 2pi, every time you find a solution if you add a multiple of 2pi to that solution you will find another solution. 1 ± , 2 β’ When the equation involves sin(x)= ± ±1, 0 then you do not need a calculator. 2 , 2 ± 3 , 2 β’ However if the sin(x)=.348 then you need to use the sin-1 on your calculator to locate the solutions. β’ The calculator gives back .355437. This is the reference angle. Knowing that sin(x) is positive an angle in the second quadrant also satisfies the equation. So take Ο β .355437 and this gives you both the 1st and 2nd quadrant angles. Now add 2kΟ to both of those to get the entire solution set. 4 cos ο± ο« 12 ο½ 8 β’ Solve this equation just like you normally would, trying to isolate the variable, understand that the variable is stuck to cosine so you are actually going to isolate cos Ο΄. 4 cos ο± ο½ ο4 cos ο± ο½ ο1 Now ask yourself, where is cos Ο΄ equal to -1. Its at Ο, and then at 3Ο, and then 5Ο So solution is Ο+2Οk 3cos ο± ο« 6 ο½ 4 3cos ο± ο½ ο2 ο2 cos ο± ο½ 3 This is a little trickier, we know that the cosine is negative so it exists in the 2nd and 3rd quadrants. But first we must find a reference angle that exists in the 1st quadrant, so neglect the negative and find cos-1(2/3). That will yield .841069. This is the 1st quadrant reference angle. Now take that reference angle and place it in the 2nd and 3rd quadrants. Ο .841069 radians or 48.1897β° Now looking for these 2 angles In order to do this, take .841069 and add it to pi to find the yellow angle, and subtract it from pi to find the red one. The same idea can be thought of using the graph of cos(Ο΄) 2 3 2 ο 3 48.2β° Slopes of lines with the use of an angle of inclination. An angle of inclination is the angle formed by a line and the horizontal (or in this case the x axis). If we look at slope as rise/run we end up with the idea of a right triangle. Rise is opposite alpha. Run is adjacent to alpha. So if we talk about the angle alpha we recognize the relationship of tangent. Ξ± run rise Thus tan(Ξ±)=rise/run Thus tan(Ξ±) = m And Ξ± = tan-1(m) m = tan(Ξ±) and Ξ± = tan-1(m) β’ Line l passes through the point (-1,3) and makes an angle of 70β° with the x-axis, find its slope to the nearest hundredths. β’ Then find the equation for line l. β’ Line l has an equation of 9x+4y=108. First where does the line cross the x-axis? What is the angle of inclination for the line? β’ β’ β’ β’ Consider the two lines. l1 ο 5x+3y=30 l2 ο 5x-2y=-10 Find the measure of the acute angle alpha that they form at their intersection. Ξ± Homework pg. 299 1-18 evens 19-24, 27, 28