# exact value using trig identities

• Apr 8th 2009, 04:26 PM
amanda0603
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!!!
• Apr 8th 2009, 04:42 PM
Prove It
Quote:

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!!!

$\displaystyle \cos^2{75^\circ} = (\cos{75^\circ})^2$

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

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

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

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

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

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

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

I'll leave you to clean that up...
• Apr 8th 2009, 04:48 PM
Prove It
Quote:

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. $\displaystyle \sec^2{15^\circ} = \frac{1}{\cos^2{15^\circ}}$

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 8 - 4\sqrt{3}$.
• Apr 8th 2009, 05:02 PM
Prove It
Quote:

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. $\displaystyle \cos{x} = \frac{1}{3}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

For this we use the difference formula...

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

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

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

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

So you were VERY close...
• Apr 8th 2009, 05:06 PM
Prove It
Quote:

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?
• Apr 8th 2009, 05:11 PM
amanda0603
IM SORRY!!!
the exact value of tan2A

THANKS SO MUCH!
• Apr 8th 2009, 05:26 PM
Prove It
Quote:

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...

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

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

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

First...

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

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

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

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

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

Now...

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

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

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

So

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

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

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

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

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