# Thread: Help with this problems!asap

1. ## Help with this problems!asap

New here so be nice

How do you find the exact value of these expressions?

sin^2 15(degrees) cot^2 75(degrees) + cos^2 15(degrees) tan^2 75(degrees)

and

1-tan(pi/6)+2csc(pi/3)

Thanks a TONNE

2. I have tryed to solve them the best I could...

Im assuming that the first one is 0 because they are the inverse of eachother.

The second one I got 2.73 by converting tan(pi/6) to sqr3/3 and csc to 2/sqr3, then solving accordingly.

Does this sound correct?

3. Originally Posted by gabrie30

How do you find the exact value of these expressions?

sin^2 15(degrees) cot^2 75(degrees) + cos^2 15(degrees) tan^2 75(degrees)
Note that:
$\displaystyle sin(15^{\circ}) = sin(45^{\circ}-30^{\circ})$
$\displaystyle cot(75^{\circ}) = cot (45^{\circ} + 30^{\circ})$

Use the sum-difference formulas to solve them. (i.e. sin(a-b) = sin(a)cos(b)-cos(a)sin(b))

1-tan(pi/6)+2csc(pi/3)
$\displaystyle tan \frac{\pi}{6}$ has exact value $\displaystyle \frac{1}{\sqrt3}$

$\displaystyle cosec \frac{\pi}{3} = \frac{1}{sin \frac{\pi}{3}}\Rightarrow sin\frac{\pi}{3}$ has exact value $\displaystyle \frac{\sqrt3}{2}$

4. Hello, gabrie30!

Welcome aboard!

Find the exact value: .$\displaystyle \sin^2\!15^o\cot^2\!75^o + \cos^2\!15^o \tan^2\!75^o$
Use the Double Angle identities:

. . $\displaystyle \begin{array}{ccccccc}\sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} & & \cos^2\!\theta &=& \dfrac{1 + \cos2\theta}{2} \\ \\[-3mm] \tan^2\!\theta &=& \dfrac{1-\cos2\theta}{1 + \cos2\theta} & & \cot^2\theta &=& \dfrac{1 + \cos2\theta}{1-\cos2\theta} \end{array}$

We have: .$\displaystyle \frac{1-\cos30^o}{2}\cdot\frac{1 + \cos150^o}{1-\cos150^o} + \frac{1+\cos30^o}{2}\cdot\frac{1 - \cos150^o}{1 + \cos150^o}$

Since $\displaystyle \cos150^o \:=\:\text{-}\cos30^p$, we have: .$\displaystyle \frac{1-\cos30^o}{2}\cdot\frac{1-\cos30^o}{1+\cos30^o} + \frac{1+\cos30^o}{2}\cdot\frac{1+\cos30^o}{1-\cos30^o}$

. . which simplfies to: .$\displaystyle \frac{(1-\cos30^o)^3 + (1+\cos30^o)^3}{2(1-\cos^230^o)} \;=\;\frac{1+3\cos^2\!30^o}{\sin^2\!30^o}$

Therefore, we have: .$\displaystyle \frac{1 + 3\left(\frac{\sqrt{3}}{2}\right)^2}{\left(\frac{1} {2}\right)^2} \;=\;\frac{1 + 3\cdot\frac{3}{4}}{\frac{1}{4}} \;=\;\boxed{13}$

$\displaystyle 1-\tan\left(\frac{\pi}{6}\right) + 2\csc\left(\frac{\pi}{3}\right)$
These are standard angles . . .
We are expected to know that: .$\displaystyle \begin{array}{ccc}\tan\left(\dfrac{\pi}{6}\right) &=& \dfrac{1}{\sqrt{3}} \\ \\[-3mm] \csc\left(\dfrac{\pi}{3}\right) &=&\dfrac{2}{\sqrt{3}} \end{array}$