Why does $\sin(90 + x) = \cos(x)$ and $\sin(90 - x) = \cos(x)$?

Question

Why are both statements below true in regards of a trigonometric circumference?

• $\sin(90 + x) = \cos(x)$
• $\sin(90 - x) = \cos(x)$

Answer 1

HINT \begin{align*} \sin(x+y) = \sin(x)\cos(y) + \sin(y)\cos(x) \end{align*}

Answer 2

The "co" in "cosine" means "complementary". The cosine of $\theta$ is the sine of the complementary angle to $\theta$, which is $90 - \theta.$ You can see this in the right triangle. If $\theta$ is one acute angle, then $90-\theta$ is the other acute angle. So $\sin(90-\theta) = $ the opposite side over the hypotenuse. But note that, in reference to $\theta$ thats the same as the adjacent side over the hypotenuse.

So that's why $\sin(90-\theta) = \cos\theta.$

Similarly, we have $\cos(90-\theta) = \sin\theta.$

Now take this last identity and replace $\theta$ by $90-\theta$. You get:

$$\cos(90-(90-\theta)) = \sin(90-\theta)$$

which is

$$\cos(\theta) = \sin(90-\theta).$$

Answer 3

The easiest way to see the given identities is to use definition of $\sin x$ and $\cos x$ using a right triangle.

Notice that if one of the acute angles in a right triangle is $x$, the other one is $90-x$

From the definition of $$\sin x = \frac {opposite}{hypotenuse}$$ and $$\cos x = \frac {adjacent}{hypotenuse}$$ you see that $$\sin x =\cos(90-x)$$ and $$\cos x =\sin(90-x)$$

Answer 4

The sum formula for sine is:-

sin(+)=sin()cos()+sin()cos()
Put x = 90 in above equation, (* angles in degrees)

sin(90 + y) = sin(90)*cos(y) + sin(y)*cos(90) .
=> sin(90 + y) = cos(y) [ as sin(90) = 1 and cos(90) = 0]
=> sin(90 + x) = cos(x) [ replace y with x]

If you replace x with -x in the above equation you get
sin(90 - x) = cos(-x)
=> sin(90 - x) = cos(x) [as cos(-x) = cos(x)]

More explanation - sin and cos are complementary to each other, that's where the name came from - sine and cosine .
Complementary angles in a triangle are x and 90-x.
So, sin(complementary angle) = cos(angle) .
sin(90-x) = cos(x) .
also cos(90-x) = sin(x)

You can get the formula for 90+x angles by replacing x with -x.

Answer 5

Consider the graph of $y= \sin(x)$.

If you slide $\sin(x)$ to the left by 90 degrees, you get $\cos(x)$. A left shift of $\sin(x)$ by 90 degrees is the same as $\sin {(x+90)}$. (One way to think about this is to ask, at what value of x does $x+90=0$? Obviously, $x=-90$, so that is the shift of the graph from the origin.) So $\sin {(x+90)} = \cos x$.

You could also flip $\sin(x)$ around the y-axis (that is, replace x with -x), then shift right by 90. A right shift by 90 changes x to x-90. That would give $\sin {(-(x-90))} = \sin {(90-x)} = \cos x$.

Answer 6

One way I have used to view the symmetry between the equations sin(x)=cos(90-x) is by visualizing two triangles:

two triangles with complementary angles

In the above image, you can find that the cosine of theta is x/hyp, while the sin of 90-theta is x/hyp.

Seeing it graphically helped he visualize how these two functions were related.

Answer 7

Here is an animation that might help explain this.

The horizontal line on the $x$-axis is the adjacent side of the right triangle whose base is the $x$-axis. The length of that side is the length of the hypotenuse times the cosine of the counter-clockwise angle between the positive $x$-axis and the hypotenuse.

The moving horizontal line is the opposite side of the right triangle whose base is the $y$-axis. The length of that side is the length of the hypotenuse times the sine of the counter-clockwise angle between the negative $y$-axis and the hypotenuse. The length of that side is also the length of the hypotenuse times the sine of the clockwise angle between the positive $y$-axis and the hypotenuse.