a Use the trigonometric ratios. For the acute angles of a general right triangle, find the sine of one angle and the cosine of the other angle.
B
b Remember that complementary angles are those angles whose measures add up to 90^(∘).
C
c Use the results from the two previous parts to write an equation that relates the sine and the cosine to an angle and its complementary angle.
A
a They are equal. Yes, the sine of one angle is equal to the cosine of the other angle.
B
b ∠ B; ∠ A.
C
c See solution.
Practice makes perfect
a Let's consider the right triangle below.
We find the sine of ∠ A by using the trigonometric ratio.
sin A = opp./hyp.
⇒
sin A = 16/34 = 8/17
Similarly, we can find the cosine of ∠ B.
cos B = adj./hyp.
⇒
cos B = 16/34 = 8/17
We can see that sin A=cos B. To know if this is true for other right triangles, let's consider the triangle below and let's focus on ∠ Q.
From the above, we have that sin Q = PR RQ. Next, on the same triangle, let's focus on ∠ R.
Then, we obtain that cos R = PR RQ. Therefore, we can state the following identity.
sin Q=cos R
In conclusion, for the acute angles of a right triangle, it is true that the sine of one angle is equal to the cosine of the other angle.
b In the right triangle below, we have that m∠ A + m∠ B = 90^(∘).
Therefore, we have that the complement of ∠ A is ∠ B. With the same reasoning, we conclude also that the complement of ∠ B is ∠ A.
c Let △ ABC be a right triangle such that ∠ C is its right angle. From Part A we can write the following result.
cos A = sin B
From Part B, we know that ∠ B is the complement of ∠ A.
cos A = sin(complement of A)
In the latter equation, we can see that it makes sense to say that the cosine is the complement of the sine.
