I am having troubles understanding how the answer key to my Pre-Calculus and Trigonometry document got to the answer it did from this task/question:

Find the exact value of the expression $$\cos(\arcsin(\frac{24}{25}))$$

At first, I tried to find the value of $\arcsin(\frac{24}{25})$ and then find the cosine of that value, but it seemed way too hard to do without a calculator. I peeked at the answer key, and I saw this:

Let $y=\arcsin(cos(\frac{24}{25})$ Then, $\sin(y)= \frac {24}{25}$ and $\cos(y)= \frac{7}{25}$

The triangle provided with the answer

I do not understand how they came to this answer or the route to it. Could someone please guide me in the right direction or show me how the document came to this answer?

  • $\begingroup$ Quick tip: use \cos instead of cos etc... to get non-italicized trigonometric functions in equation mode. $\endgroup$ – FraGrechi Aug 17 '16 at 17:20
  • $\begingroup$ It's complicate to do easier ! $\cos(y)=\frac{7}{25}$ and $y=\arcsin(24/25)$. The claim follow... these are just the definition of sinus, cosinus... $\endgroup$ – Surb Aug 17 '16 at 17:22

Let's break it down like this.

Remember $\sin$ and $\arcsin$ are inverse operations. So if we take the $\sin$ of some angle in a right triangle, and our answer is 24/25, then that means the opposite side of that angle is 24, and the hypotenuse is 25. And we do the inverse with $\arcsin$. If we do $\arcsin(24/25)$, the answer will be the size of an angle in a right triangle whose opposite side is 24, and whose hypotenuse is 25.

In summary, $\sin$ takes angle and gives opp/hyp, and $\arcsin$ takes opp/hyp, and gives angle.

So $\arcsin(24/25)$ gives you the angle. Now taking the $\cos$ of that angle means we have to find the adj/hyp ratio. We already know that the hypotenuse is 25, so we just need to fill in the adj part.

Since we know the opposite side and the hypotenuse, we can just use the pythagorean theorem to find the adjacent side.

$$ 25^2 = 24^2 + \text{adj}^2 $$ $$ \text{adj} = 7 $$ And so the $\cos$ of the angle given by $\arcsin(24/25)$ is 7/25.

  • $\begingroup$ Thanks! I forgot that "arcsin takes opp/hyp, and gives angle" so nothing made sense. $\endgroup$ – Blake Aug 17 '16 at 17:56

Let $\arcsin \frac{24}{25}=\alpha$.

Find $\cos \alpha$.

If $\arcsin \frac{24}{25}=\alpha$, than $$\sin \left(\arcsin \frac{24}{25}\right)=\sin \alpha$$ $$\sin \alpha=\frac{24}{25}$$ Then $$\cos \alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}$$

Answer: $\cos (\arcsin \frac{24}{25})=\frac{7}{25}$


The definition of $\sin y$ is that it is the length of the side opposite the angle ($24$ in the picture) divided by the length of the hypotenuse ($25$ in the picture).

So the thing to do is to draw a right-angled triangle with one side $24$, hypotenuse $25$, and work out the length of the remaining side. The square of that length will be $25^2-24^2$, which is $49$, which is $7^2$. So the remaining side is $7$.

The definition of $\cos y$ is that it is that short side divided by the hypotenuse - so, $\frac{7}{25}$.


The basic definition of "sine" is "opposite side divided by hypotenuse" so they have drawn a right triangle with angle (call it $\theta$) on the left, hypotenuse of length 25 and "opposite side" (the vertical line on the right) of length 24. So $\sin(\theta)= \frac{24}{25}$ and $\theta= \arcsin\left(\frac{24}{25}\right)$.

By the Pythagorean theorem, then, $x^2+ 24^2= x^2+ 576= 25^2= 625$ where "x" is the length of the "near side", the horizontal line at the bottom of the picture. So $x^2= 625- 576= 49$. Since the length of a side must be positive, the length of the "near side" is $\sqrt{49}= 7$ as shown in the picture.

"cosine" is defined as "near side over hypotenuse" we have $cos\left(sin\left(\frac{24}{25}\right)\right)= cos(\theta)= \frac{7}{25}$.

A more "trigonometric" method of getting that result is to use the fact that $sin^2(\theta)+ cos^2(\theta)= 1$. Setting $\theta= arcsin\left(\frac{24}{25}\right)$ we have $sin(\theta)= \frac{24}{25}$ and $sin^2(\theta)+ cos^2(\theta)= \left(\frac{24}{25}\right)^2+ cos^2(\theta)= 1$ so $cos^2(\theta)= 1- \frac{576}{625}= \frac{625- 576}{625}= \frac{49}{625}$ and then $cos(\theta)= \sqrt{\frac{49}{625}}= \frac{7}{25}$.


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