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Illustrative
Mathematics
G-SRT Trigonometric Function
Values
Alignments to Content Standards: G-SRT.C.7
Task
a. For which acute angles a is sin a
= cos a? Explain.
b. For which acute angles a is sin a
= cos 2a? Explain.
IM Commentary
The goal of this task is to explore the relationship between sine and cosine of
complementary angles for special benchmark angles. Students should recognize that
sin 45 = cos 45 =
√2
and should also recognize that
2
sin 60 = cos 30 =
√3
. On the
2
other hand, in order to explain why they have found all solutions to the equations,
students will need to use additional knowledge about the sine and cosine functions.
This may include using the relationships sin x = cos (90 − x) and cos x = sin (90 − x)
or they may also use the fact that over the interval 0 ≤ x ≤ 90 the function sin x is
increasing and the function cos x is decreasing.
Edit this solution
Solution
a. One way to do this problem uses the definition of sine and cosine in terms of a right
triangle with an acute angle of measure x: sin x is the length of the side opposite the
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Illustrative
Mathematics
angle divided by the length of the hypotenuse while cos x is the length of the side
adjacent to the angle divided by the length of the hypotenuse. If sin x = cos x this
means that the side opposite the angle and the side adjacent to the angle are
congruent. In this case, the right triangle is a right isosceles triangle and both acute
angles measure 45 ∘ .
Another way to do this problem uses the relationship between trigonometric functions
and coordinates of points on the unit circle. Using degrees to measure angles, for each
acute angle 0 < a < 90 we can make an angle with the positive x-axis as one ray and
whose counterclockwise measure is a as pictured below:
Also shown in the picture is the point of intersection (cos a, sin a) of the angle's ray
with the unit circle. The only point in the first quadrant of the unit circle whose x and y
coordinates are equal is the point ( √
of equations x = y and x 2 + y2
have x = 45 if sin x = cos x.
2 √2
, 2 ), which can be shown by solving the pair
2
= 1. This makes a 45 degree angle and so we must
sin 45 = cos 45 =
This give us one angle, namely x = 45, with sin x = cos x. But sin x is an increasing
function for 0 ≤ x ≤ 90 while cosine is a decreasing function on this domain. This
means that for any 0 ≤ x < 45, sin x < cos x while for any 45 < x ≤ 90 we have
sin x > cos x. So sin x = cos x only when x = 45.
A third way we can solve this problem using the knowledge that
√2
.
2
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Mathematics
b. For this problem, we can use the second line of reasoning for part (a). We know that
sin x is non-negative and increases for 0 ≤ x ≤ 90 while cos 2x decreases for
0 ≤ x ≤ 90. This means that sin x can take the same value as cos 2x exactly once.
Moreover, we must have sin x = cos 2x for some acute angle because sin 0 < 2 cos 0
while sin 90 > 2 cos 90. Using knowledge of special angle values of trigonometric
functions we have sin 30 = cos 60 = 1 . So sin x = cos 2x when x = 30. Alternatively
we can write cos 2x
when x = 30.
2
= sin (90 − 2x) and then check that sin x = sin (90 − 2x) exactly
An alternate solution to part (b) is to use trig identities rather than noticing special
values of trig functions. Then using, for example, the identity cos(2a) = 1 − 2 sin2 (a),
we are left with the equation sin(a) = 1 − 2 sin2 (a). Solving for sin(a) gives
sin(a) =
−1 ± √‾1‾‾‾‾‾‾‾‾‾‾
− 4(2)(−1)‾
4
= −1 or 1/2.
Since an angle a with sin(a) = −1 represents a non-acute angle, the unique solution
we are looking for has sin(a) = 1 , and so a = sin−1 ( 1 ) (which of course, recovers the
2
2
value of 30 degrees, or π/6 radians, from the other solution).
G-SRT Trigonometric Function Values
Typeset May 4, 2016 at 21:25:23. Licensed by Illustrative Mathematics under a
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