Trigonometric Function Values

Task

1. For which acute angles $a$ is $\sin{a} = \cos{a}$? Explain.
2. 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 $\sin{45} = \cos{45} = \frac{\sqrt{2}}{2}$ and should also recognize that $\sin{60} = \cos{30} = \frac{\sqrt{3}}{2}$. On the 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 $0 \leq x \leq 90$ the function $\sin{x}$ is increasing and the function $\cos{x}$ is decreasing. 

Solution

1. 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 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$^\circ$.

   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 $0 \lt a \lt 90$ 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 $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$, which can be shown by solving the pair of equations $x=y$ and $x^2+y^2=1$. This makes a 45 degree angle and so we must have $x = 45$ if $\sin{x}=\cos{x}$.

   A third way we can solve this problem using the knowledge that $\sin{45} = \cos{45} = \frac{\sqrt{2}}{2}$. This give us one angle, namely $x = 45$, with $\sin{x} = \cos{x}$. But $\sin{x}$ is an increasing function for $0 \leq x \leq 90$ while cosine is a decreasing function on this domain. This means that for any $0 \leq x \lt 45$, $\sin{x} \lt \cos{x}$ while for any $45 \lt x \leq 90$ we have $\sin{x} \gt \cos{x}$. So $\sin{x} = \cos{x}$ only when $x = 45$.

    

2. 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 $0 \leq x \leq 90$ while $\cos{2x}$ decreases for $0 \leq x \leq 90$. 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 $\sin{0} \lt 2\cos{0}$ while $\sin{90} \gt 2\cos{90}$. Using knowledge of special angle values of trigonometric functions we have $\sin{30} = \cos{60} = \frac{1}{2}$. So $\sin{x} = \cos{2x}$ when $x = 30$. Alternatively we can write $\cos{2x} = \sin{(90-2x)}$ and then check that $\sin{x} = \sin{(90-2x)}$ exactly when $x = 30.$

   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 $\cos(2a)=1-2\sin^2(a)$, we are left with the equation $\sin(a)=1-2\sin^2(a)$. Solving for $\sin(a)$ gives $$\sin(a)=\frac{-1\pm \sqrt{1-4(2)(-1)}}{4}=-1\text{ or }1/2.$$ Since an angle $a$ with $\sin(a)=-1$ represents a non-acute angle, the unique solution we are looking for has $\sin(a)=\frac{1}{2}$, and so $a=\sin^{-1}(\frac{1}{2})$ (which of course, recovers the value of 30 degrees, or $\pi/6$ radians, from the other solution).