Thread: Trigonometric Identities problem from my pre-calc class that is driving me insane

1. Trigonometric Identities problem from my pre-calc class that is driving me insane

So the problem is a verify the equation problem where the left side has to be made equal to the right : sin-1 * (4/5) - cos-1 * (12/13) = cos-1* (56/65)

- My confusion come from two main things
1. I've never done a problem with negative 1 exponents and I can't find an example of one in my book
2. I've done a hundred of these verify problems but I've never come across one that is sin - cos = cos

Also my first time posting so I may have done something wring

2. Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

The "negative one exponent" indicates the inverse function. $\displaystyle sin^{-1}(x)$ and $\displaystyle cos^{-1}(x)$ are also often called "arcsin(x)" and "arcos(x)". If cos(x)= y then $\displaystyle cos^{-1}(y)= x$. I find it very hard to believe that your textbook gave you a problem like this without ever having defined them!

3. Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

Your Q. asks you to "verify". Note that in mathematics, that is not the same as "prove": "verify" has the same meaning as "validate".

On those grounds, a quick use of a calculator reveals that:
sin-1 * (4/5) = 53.130°,
cos-1 * (12/13) = 22.620°
Difference = 30.510°
And that cos-1* (56/65) = 30.510°,
Thus completing the verification - as required.

Aside: Sorry about the "sin-1" typing. I did a cut-and-paste from the OP's Q., but that operation, strangely enough, fails here.

Al.

4. Re: Trigonometric Identities problem from my pre-calc class that is driving me insane

an inverse trig function value is an angle ...

$\sin{x} = y \implies x = \sin^{-1}{y}$

let $\theta = \sin^{-1}\left(\dfrac{4}{5}\right)$

let $\phi = \cos^{-1}\left(\dfrac{12}{13}\right)$

note both angles have terminal sides in quadrant I $\implies \cos{\theta} = \dfrac{3}{5}$ and $\sin{\phi} = \dfrac{5}{13}$

$\cos(\theta - \phi) = \cos{\theta}\cos{\phi} + \sin{\theta}\sin{\phi}$

$\cos(\theta - \phi) = \dfrac{3}{5} \cdot \dfrac{12}{13} + \dfrac{4}{5} \cdot \dfrac{5}{13}$

$\cos(\theta - \phi) = \dfrac{36}{65} + \dfrac{20}{65}$

$\cos(\theta - \phi) = \dfrac{56}{65}$

$\cos^{-1}[\cos(\theta - \phi)] = \cos^{-1}\left(\dfrac{56}{65}\right)$

$\theta - \phi = \cos^{-1}\left(\dfrac{56}{65}\right)$