The concept behind sin(π/2-Φ)=cos(Φ)

**Oiler** · January 11th 2011, 08:50 AM

Hey, I am having difficulty understanding the concept behind sin(π/2-Φ)=cos(Φ). I know it means the angle is in the first quadrant and that sine is positive in that quadrant. How does it equal cos(Φ)? How can i prove this using the triangles or unit circle. Thanks

**dwsmith** · January 11th 2011, 09:07 AM

The x and y axis are 90 degrees apart. So by adding or subtracting 90, you are moving to the other axis. You can use the sum identity.

$\sin{\left(\frac{\pi}{2}-\theta\right)}=\sin{\frac{\pi}{2}\cos{\theta}-\cos{\frac{\pi}{2}\sin{\theta}$

What does this simplify to?

**TheEmptySet** · January 11th 2011, 09:50 AM

Originally Posted by Oiler

Hey, I am having difficulty understanding the concept behind sin(π/2-Φ)=cos(Φ). I know it means the angle is in the first quadrant and that sine is positive in that quadrant. How does it equal cos(Φ)? How can i prove this using the triangles or unit circle. Thanks

The concept behind sin(π/2-Φ)=cos(Φ)-capture.jpg

If you look at this picture you can see that
$\displaystyle \cos(\theta)=\frac{b}{\text{hypotenuse}}$ and also that
$\displaystyle \sin\left( \frac{\pi}{2}-\theta\right)=\frac{b}{\text{hypotenuse}}$

You can draw pictures to verify this in other quadrants as well.

**skeeter** · January 11th 2011, 12:10 PM

$\displaystyle \frac{\pi}{2} - \phi$ is the complement of the angle $\phi$ ... cosine is the sine of an angle's complement.

**Oiler** · January 13th 2011, 12:18 PM

Using a similar picture, how can i go ahead to show that $\left(\frac{\pi}{2}+\theta \right ) = \cos\theta$ . Thanks.

**dwsmith** · January 13th 2011, 12:20 PM

Originally Posted by dwsmith

The x and y axis are 90 degrees apart. So by adding or subtracting 90, you are moving to the other axis. You can use the sum identity.

$\sin{\left(\frac{\pi}{2}-\theta\right)}=\sin{\frac{\pi}{2}\cos{\theta}-\cos{\frac{\pi}{2}\sin{\theta}$

What does this simplify to?

$\sin{\left(\frac{\pi}{2}+\theta\right)}=\sin{\frac {\pi}{2}\cos{\theta}+\cos{\frac{\pi}{2}\sin{\theta }$

**Oiler** · January 13th 2011, 01:14 PM

I am able to use the formula and correctly obtain $\cos\theta$ . I Have been trying to perhaps use the unit circle or a triangle to prove the equation.

**dwsmith** · January 13th 2011, 01:16 PM

$\sin{\frac{\pi}{2}}=1 \ \ \ \cos{\frac{\pi}{2}}=0$

$\sin{\left(\frac{\pi}{2}+\theta\right)}=\sin{\frac {\pi}{2}\cos{\theta}+\cos{\frac{\pi}{2}\sin{\theta }=1*\cos{\theta}+0*\sin{\theta}=\cos{\theta}$

**Oiler** · January 14th 2011, 02:56 PM

dwsmith, I understand how that formula works, Just not pictorially like how TheEmptySet answered the first question.

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