Question Video: Finding Equivalent Expressions Using the Cofunction Identity for Sine and Cosine | Nagwa Question Video: Finding Equivalent Expressions Using the Cofunction Identity for Sine and Cosine | Nagwa

# Question Video: Finding Equivalent Expressions Using the Cofunction Identity for Sine and Cosine Mathematics • First Year of Secondary School

## Join Nagwa Classes

Which of the following is equal to sin 𝜃? [A] sin ((3𝜋/2) + 𝜃) [B] cos ((3𝜋/2) + 𝜃) [C] sin ((𝜋/2) + 𝜃) [D] cos ((𝜋/2) + 𝜃)

05:11

### Video Transcript

Which of the following is equal to sin of 𝜃? (A) sin of three 𝜋 over two plus 𝜃. (B) cos of three 𝜋 over two plus 𝜃. (C) sin of 𝜋 over two plus 𝜃. Or (D) cos of 𝜋 over two plus 𝜃.

We begin by sketching an arbitrary angle 𝜃 in standard position on an 𝑥𝑦-coordinate plane. Then, we sketch a unit circle, centered at the origin.

We recall that an angle 𝜃 in standard position will intersect the unit circle at a point with the 𝑥-coordinate cos of 𝜃 and the 𝑦-coordinate sin of 𝜃. We then sketch the reference triangle for angle 𝜃 along the positive side of the 𝑥-axis. We know that the hypotenuse of a reference triangle is a radius of the unit circle and thus has a length of one. We also note that the side of our reference triangle, which runs along the positive 𝑥-axis, corresponds to cos of 𝜃.

In the first and fourth quadrants, the 𝑥-coordinate is always positive. But in the second or third quadrant, 𝑥 is negative. The side of the reference triangle opposite 𝜃 corresponds to the 𝑦-coordinate or sin of 𝜃. The 𝑦-coordinate is positive in the first and second quadrants but negative in the third and fourth quadrants.

Now that we have reviewed where sine is found in relation to angle 𝜃 and its reference triangle, we will sketch a congruent triangle for each of the other two angles mentioned in this question, three 𝜋 over two plus 𝜃 and 𝜋 over two plus 𝜃.

To locate the angle three 𝜋 over two plus 𝜃, we locate the angle three 𝜋 over two and rotate 𝜃 in the positive direction. Three 𝜋 over two is located on the negative 𝑦-axis. From there, we rotate counterclockwise by 𝜃. This angle intersects the unit circle at the coordinate point cos of three 𝜋 over two plus 𝜃, sin of three 𝜋 over two plus 𝜃.

Next, we locate 𝜋 over two plus 𝜃 by rotating 𝜃 past the positive 𝑦-axis in the counterclockwise direction. This angle intersects the unit circle at the coordinate point cos of 𝜋 over two plus 𝜃, sin of 𝜋 over two plus 𝜃. Using the angle 𝜋 over two plus 𝜃, we can sketch a right triangle that is congruent to our original reference triangle. But this line will be connected to the 𝑦-axis.

The exact location of this new right triangle is found by simply rotating the original reference triangle 90 degrees or 𝜋 over two radians. So it has the same dimensions as the original reference triangle, which had a longer side of cos of 𝜃 and a shorter side of sin of 𝜃. Because 𝜋 over two plus 𝜃 is in the second quadrant, the 𝑥-coordinate will be negative. Since the shorter side of the new triangle now represents the 𝑥-value, it will be negative sin of 𝜃. And the other side of the triangle represents the 𝑦-value, which will still be positive cos of 𝜃. So our coordinate point can be written in terms of 𝜃 as negative sin of 𝜃, cos of 𝜃.

We have just shown that cos of 𝜋 over two plus 𝜃 equals negative sin of 𝜃 and sin of 𝜋 over two plus 𝜃 equals cos of 𝜃. Thus, we eliminate options (C) and (D) because neither expression equals sin of 𝜃.

Now we return to the angle three 𝜋 over two plus 𝜃. Here, we will sketch another triangle congruent to our reference triangle. This new triangle is placed along the negative 𝑦-axis. In quadrant four, the 𝑦-coordinate will be negative, but the 𝑥-coordinate is once again positive, as it was in the first quadrant. We find the coordinate point in terms of 𝜃 to be sin of 𝜃, negative cos of 𝜃. This means that cos of three 𝜋 over two plus 𝜃 equals sin of 𝜃 and sin of three 𝜋 over two plus 𝜃 equals negative cos of 𝜃.

In conclusion, we have shown that option (B) is the expression that equals sin of 𝜃. This is because on the unit circle, the 𝑥-coordinate of the point given by the angle three 𝜋 over two plus 𝜃 is the same as the 𝑦-coordinate given by the angle 𝜃.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions