How to solve $\cos(\sin^{-1}(-3/5))$?
I'm stuck with question $$\cos\Bigl(\sin^{-1}\Bigl(-\frac35\Bigr)\Bigr)$$ I looked for the answer in the book and it is $\frac45$
I tried solving it using the formula $\sin^2x+\cos^2x=1$and I also got $\frac45$ as the answer but I got it by inputting the value of $\cos x$ in the given question in place of$$ \sin^{-1}\Bigl(-\frac35\Bigr)$$But I cant see any logic there.
Please explain it to me.
$\endgroup$6 Answers
$\begingroup$Hint:$$\cos(\arcsin(x))=\sqrt{1-x^2},$$ for all $x\in (-\pi/2,\pi/2)$.
$\endgroup$ 3 $\begingroup$You can do this purely with (trigonometric) formulas, but here's a more geometric approach.
Since $\sin^{-1}(-a)=-\sin^{-1}(a)$ and $\cos(-a)=\cos(a)$, you have:$$\cos\left(\sin^{-1}\left(-\tfrac{3}{5}\right)\right)=\cos\left(\sin^{-1}\left(\tfrac{3}{5}\right)\right)$$Now imagine (or draw!) a right triangle with sides 3, 4 and 5:
- $\sin^{-1}\left(\tfrac{3}{5}\right)$ corresponds to the angle with opposite side 3;
- the cosine of this angle is the ratio of the adjacent side (4) and the hypotenuse (5).
Actually, for better understanding, you can translate this to English. For this particular expression$$\cos\Bigl(\sin^{-1}\Bigl(-\frac35\Bigr)\Bigr)$$what it asks is
What is the cosine of an angle whose sine value is $-\frac{3}{5}$?
Now, there are two things to be careful here. First, we need to find an angle whose sine value is $-\frac{3}{5}$. Second, we are asked to find the cosine value of this angle.
From here, I refer to StackTD's answer since I find solving it geometrically more beneficial.
$\endgroup$ $\begingroup$$\displaystyle \sin^{-1}\left(-\dfrac 35 \right)$ is the angle with its sine ratio equal to $-\dfrac35$. Note that angles in the first and the second quadrants have positive sine ratios and angles in the third and the fourth quadrants have negative sine ratios. So, $\displaystyle \sin^{-1}\left(-\dfrac 35 \right)$ must be an angle in the third or the fourth quadrant. We don't want $\sin^{-1}$ to have multiple values and therefore we restrict its range to $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}2\right]$, with $\sin^{-1}$ of a positive number not greater than $1$ has the value in $\displaystyle \left(0, \frac{\pi}2\right]$ and that of a negative number not less than $-1$ has the value in $\displaystyle \left[-\frac{\pi}{2}, 0\right)$ (Also, we have $\sin^{-1}0=0$). This is the way we define the inverse sine function.
So if $\displaystyle \theta = \sin^{-1}\left(-\dfrac 35 \right)$, then $\theta$ is an angle in $\displaystyle \left[-\frac{\pi}{2}, 0\right)$ such that $\displaystyle \sin\theta=-\dfrac35$. $\theta$ is an angle in the fourth quadrant and hence $\cos\theta\ge0$.
As $\sin^2\theta+\cos^2\theta=1$, $\cos^2\theta=\displaystyle 1-\left(-\dfrac35\right)^2=\frac{16}{25}$. $\cos\theta\ge0$ implies that $\cos\theta=\dfrac45$.
$\endgroup$ $\begingroup$Let $\sin^{-1}\left(-\frac 35\right)=\alpha$ so that $\sin \alpha =-\frac 35$
Then $\sin^2\alpha +\cos ^2 \alpha =1$ and $\cos^2 \alpha = \frac {16}{25}$
I think this is essentially want you did - or similar to it, and all you need then is some care about which square root you want - which in reality may depend on context, but when no context is stated there is a convention for the principal value which others have explained.
I just wanted to highlight that giving a name like $\alpha$ to a complicated expression can sometimes help you to see what is going on. It is often useful to use a name other than commonly used variable names like $x$ or $n$, because his avoids using the same symbol to mean different things.
$\endgroup$ $\begingroup$To find $\cos(\arcsin(-3/5))$, since $\sin(x)=-3/5 $, $x=\arcsin(-3/5)$. Then$\cos(x)=\pm \sqrt{1-(3/5)^2 }$, since $\cos^2 x+\sin^2 x=1$, i.e. $\cos(x) = \pm 4/5$.
$\endgroup$
