# Thread: exact value using trig identities

1. ## exact value using trig identities

1.
cos^2(75degrees)
2.
sec^2(15degrees)

also
FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
sin(x-B)

I got (2(squareroot)6-1)/6 but Im not 100% sure.

Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.

Any help is gratefully appreciated!!!

2. Originally Posted by amanda0603

1.
cos^2(75degrees)
2.
sec^2(15degrees)

also
FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
sin(x-B)

I got (2(squareroot)6-1)/6 but Im not 100% sure.

Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.

Any help is gratefully appreciated!!!
$\cos^2{75^\circ} = (\cos{75^\circ})^2$

$= [\cos{(60^\circ + 15^\circ)}]^2$.

Now using the sum identity for cosine, $\cos{(a + b)} = \cos{a}\cos{b} - \sin{a}\sin{b}$, this becomes...

$(\cos{60^\circ}\cos{15^\circ} - \sin{60^\circ}\sin{15^\circ})^2$

$= \left(\frac{1}{2}\cos{15^\circ} - \frac{\sqrt{3}}{2}\sin{15^\circ}\right)^2$.

Now you need the half angle identities, $\sin{\frac{\theta}{2}} = \pm\sqrt{\frac{1 - \cos{\theta}}{2}}$ and $\cos{\frac{\theta}{2}} = \pm\sqrt{\frac{1 + \cos{\theta}}{2}}$. Notice that $15 = \frac{30}{2}$.

$\left(\frac{1}{2}\cos{15^\circ} - \frac{\sqrt{3}}{2}\sin{15^\circ}\right)^2 = \left(\frac{1}{2}\sqrt{\frac{1 + \cos{30^\circ}}{2}} - \frac{\sqrt{3}}{2}\sqrt{\frac{1 - \cos{30^\circ}}{2}}\right)^2$

$= \left(\frac{1}{2}\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} - \frac{\sqrt{3}}{2}\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}\right)^2$.

I'll leave you to clean that up...

3. Originally Posted by amanda0603

1.
cos^2(75degrees)
2.
sec^2(15degrees)

also
FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
sin(x-B)

I got (2(squareroot)6-1)/6 but Im not 100% sure.

Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.

Any help is gratefully appreciated!!!
2. $\sec^2{15^\circ} = \frac{1}{\cos^2{15^\circ}}$

$= \frac{1}{(\cos{15^\circ})^2}$

Now use the half angle identity $\cos{\frac{\theta}{2}} = \pm \sqrt{\frac{1 + \cos{\theta}}{2}}$

$= \frac{1}{\left(\sqrt{\frac{1 + \cos{30^\circ}}{2}}\right)^2}$

$= \frac{1}{\left(\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}\right)^2}$

$= \frac{1}{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

$= \frac{2}{1 + \frac{\sqrt{3}}{2}}$

(and if you want to rationalise the denominator...)

$= \frac{2}{\frac{2 + \sqrt{3}}{2}}$

$= \frac{4}{2 + \sqrt{3}}$

$= \frac{4(2 - \sqrt{3})}{4 - 3}$

$= 8 - 4\sqrt{3}$.

4. Originally Posted by amanda0603

1.
cos^2(75degrees)
2.
sec^2(15degrees)

also
FInd the exact value given that cosx=1/3 with x in quad I and sinB= -1/2 with B in quad IV.
sin(x-B)

I got (2(squareroot)6-1)/6 but Im not 100% sure.

Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.

Any help is gratefully appreciated!!!
3. $\cos{x} = \frac{1}{3}$

You need to use the Pythagorean Identity, $\sin^2{x} + \cos^2{x} = 1$.

$\sin^2{x} + \cos^2{x} = 1$

$\sin^2{x} + \left(\frac{1}{3}\right)^2 = 1$

$\sin^2{x} + \frac{1}{9} = 1$

$\sin^2{x} = \frac{8}{9}$

$\sin{x} = \frac{\sqrt{8}}{3}$

$\sin{x} = \frac{2\sqrt{2}}{3}$ (We know that $\sin{x} > 0$ since x is in the first quadrant).

We also know $\sin{B} = -\frac{1}{2}$. Once again, use the Pythagorean Identity.

$\sin^2{B} + \cos^2{B} = 1$

$\left(-\frac{1}{2}\right)^2 + \cos^2{B} = 1$

$\frac{1}{4} + \cos^2{B} = 1$

$\cos^2{B} = \frac{3}{4}$

$\cos{B} = \frac{\sqrt{3}}{2}$ (we know that $\cos{B} > 0$ because B is in the fourth quadrant.

Now we have $\sin{x}, \cos{x}, \sin{B}, \cos{B}$.

We were trying to find $\sin{(x - B)}$.

For this we use the difference formula...

$\sin{(x - B)} = \sin{x}\cos{B} - \cos{x}\sin{B}$

$= \frac{2\sqrt{2}}{3}\left(\frac{\sqrt{3}}{2}\right) - \frac{1}{3}\left(-\frac{1}{2}\right)$

$= \frac{\sqrt{6}}{3} + \frac{1}{6}$

$= \frac{2\sqrt{6} + 1}{6}$.

So you were VERY close...

5. Originally Posted by amanda0603

Find the exact value given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.
Find the exact value of what?

6. IM SORRY!!!
the exact value of tan2A

THANKS SO MUCH!

7. Originally Posted by amanda0603

Find the exact value of tan2A given that sinA= -4/5 with A in quad IV, tanB= 7/24 with B in quad III, and cosC= -5/13 with C in quad II.
You need to use the following identities...

$\sin^2{x} + \cos^2{x} = 1$,

$\tan{x} = \frac{\sin{x}}{\cos{x}}$

$\tan{2x} = \frac{2\tan{x}}{1 - \tan^2{x}}$.

First...

$\sin^2{A} + \cos^2{A} = 1$

$\left(-\frac{4}{5}\right)^2 + \cos^2{A} = 1$

$\frac{16}{25} + \cos^2{A} = 1$

$\cos^2{A} = \frac{9}{25}$

$\cos{A} = \frac{3}{5}$ (We know that $\cos{A} > 0$ because A is in quadrant 4.

Now...

$\tan{A} = \frac{\sin{A}}{\cos{A}}$

$= \frac{-\frac{4}{5}}{\frac{3}{5}}$

$= -\frac{4}{3}$.

So

$\tan{2A} = \frac{2\tan{A}}{1 - \tan^2{A}}$

$= \frac{2\left(-\frac{4}{3}\right)}{1 - \left(-\frac{4}{3}\right)^2}$

$= \frac{-\frac{8}{3}}{1 - \frac{16}{9}}$

$= \frac{-\frac{8}{3}}{-\frac{7}{9}}$

$= \frac{24}{7}$.