# Thread: The concept behind sin(π/2-Φ)=cos(Φ)

1. ## The concept behind sin(π/2-Φ)=cos(Φ)

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

2. 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.

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

What does this simplify to?

3. 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

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

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

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

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

6. 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.

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

What does this simplify to?
$\displaystyle \sin{\left(\frac{\pi}{2}+\theta\right)}=\sin{\frac {\pi}{2}\cos{\theta}+\cos{\frac{\pi}{2}\sin{\theta }$

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

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

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

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

10. ## Re: The concept behind sin(π/2-Φ)=cos(Φ)

I do not think you can use formulas lie sin(x+y) or cos(x+y) to prove that sin(x)=cos(pi/2-x) or cos(x)=sin(pi/2-x). These formulas are more advanced and should be addressed after more elementary conversions from sin to cos.
Also formulas sin(x) = -sin(pi+x), sin(x) = sin(pi-x), cos(x) = -cos(pi+x), cos(x) = -cos(pi-x) are even more elementary and follow quite easily from the definition of the sin and cos on a unit circle.

On my site unizor.com I suggested the following proof of conversion of sin into cos and back:

Theorem:
sin(φ) = cos(π/2 − φ)
cos(φ) = sin(π/2 − φ)

Proof:
First of all, let's prove this identity for any acute angle φ. In this case the general definition of functions sin and cos through a unit circle is equivalent to a definition based on the ratio between catheti and a hypotenuse in a right triangle with this acute angle. The value of sin of an acute angle in a right triangle is a ratio between an opposite cathetus and a hypotenuse, while the value of cos of this acute angle is a ratio between an adjacent cathetus and a hypotenuse.
But, the opposite acute angle to our acute angle φ in a right triangle equals to (π/2 − φ). The cathetus opposite to φ is adjacent to that angle (π/2 − φ) and the cathetus adjacent to φ is opposite to that angle (π/2 − φ).
Therefore, sin φ = cos(π/2 − φ) and cos φ = sin(π/2 − φ).
Let's consider a general case now.
We know that
sin 0 = 0, cos π/2 = 0.
Therefore, for φ=0 the identity
sin φ = cos (π/2 − φ) is true.
Also, for φ=π/2 the identity
cos φ = sin (π/2 − φ) is true.
Similarly,
sin π/2 = 1, cos 0 = 1.
Therefore, for φ=π/2 the identity
sin φ = cos (π/2 − φ) is true.
Also, for φ=0 the identity
cos φ = sin (π/2 − φ) is true.
From all the above we can conclude that for any 0≤φ≤π/2 (that is, for any angle φ in the first quadrant)
sin φ = cos (π/2 − φ) is true,
cos φ = sin (π/2 − φ) is true.
Here is a logic that can help to prove these identities for any angle.
For an angle φ in the second quadrant (that is, greater than π/2 but not exceeding π) we can say that an angle χ = π − φ lies in the first quadrant. Let's use the equality φ = π − χ and identities proved for an angle χ in the first quadrant.
sin φ = sin (π − χ) =
= sin χ = cos (π/2 − χ) =
= cos (π/2 − (π − φ)) =
= cos (φ − π/2) =
= cos (π/2 − φ).
cos φ = cos (π − χ) =
= −cos χ = −sin (π/2 − χ) =
= −sin (π/2 − (π − φ)) =
= −sin (φ − π/2) =
= −(−sin (π/2 − φ)) =
= sin (π/2 − φ).
This concludes the proof of our identity for angles in the second quadrant. For an angle φ in the third or forth quadrant we will use analogous approach. In this case an angle χ = φ − π lies in the first or second quadrant, where we have already proved the identity.
Therefore,
sin φ = sin (χ + π) =
= −sin χ = −cos (π/2 − χ) =
= −cos (π/2 − (φ − π)) =
= −cos (3π/2 − φ) =
= −(−cos (π/2 − φ)) =
= cos (π/2 − φ).
cos φ = cos (χ + π) =
= −cos χ = −sin (π/2 − χ) =
= −sin (π/2 − (φ − π)) =
= −sin (3π/2 − φ) =
= −(−sin (π/2 − φ)) =
= sin (π/2 − φ).
The identity is proved for all angles.

11. ## Re: The concept behind sin(π/2-Φ)=cos(Φ)

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

12. ## Re: The concept behind sin(π/2-Φ)=cos(Φ)

That's absolutely right, but works only for acute angles. I also started from this proof in my response above. But then you need some ingenuity to extend the proof for any angles. The equality is true for any angle, not necessarily less then 90 degrees.