1. ## Trignometric Identity Questions

I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

If somebody could help me with these questions that would be great.

1. Simplify:
$\displaystyle cos2A/(cosA+sinA)$

2. Prove:
$\displaystyle sin^2(2)A=sin^2A-sin^4A$

3. $\displaystyle sqrt(2)cos(A-(pi/4))=cosA+sinA$

4. Solve for all possible values of X:
$\displaystyle -5sin(15(x-17degrees))-3=0$

excuse my lack of math typing skills

Originally Posted by Liam
I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

If somebody could help me with these questions that would be great.

1. Simplify:
$\displaystyle cos2A/(cosA+sinA)$
$\displaystyle \frac{\cos 2A}{\cos A + \sin A}$

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

$\displaystyle =\frac{(\cos A + \sin A)(\cos A - \sin A)}{\cos A + \sin A}$

$\displaystyle =\cos A - \sin A$

we used the formula here, $\displaystyle \cos 2A=\cos^2 A - \sin^2 A$

and, $\displaystyle a^2 - b^2 = (a+b)(a-b)$

3. Originally Posted by Liam
I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

If somebody could help me with these questions that would be great.
2. Prove:
$\displaystyle sin^2(2)A=sin^2A-sin^4A$
Right Side $\displaystyle = \sin^2 A(1 - \sin^2 A)$

$\displaystyle = \sin^2 A \cos^2 A$

$\displaystyle = \frac{4\sin^2 A \cos^2 A}{4}$

$\displaystyle = \frac{(2\sin A \cos A)^2}{4}$

$\displaystyle = \frac{\sin^2 2A}{4}$

we used the formula here, $\displaystyle 2 \sin A \cos A = \sin 2A$

4. Originally Posted by Liam
I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

If somebody could help me with these questions that would be great.
3. $\displaystyle sqrt(2)cos(A-(pi/4))=cosA+sinA$
$\displaystyle =\sqrt{2} \cos \left(A - \frac{\pi}{4}\right)$

$\displaystyle = \sqrt{2}\left[\cos A .\cos \left(\frac{\pi}{4}\right) + \sin A .\sin \left(\frac{\pi}{4}\right)\right]$

$\displaystyle = \sqrt{2} \left( \cos A .\frac{1}{\sqrt{2}}+\sin A. \frac{1}{\sqrt{2}}\right)$

$\displaystyle = \cos A + \sin A$

Formula used : $\displaystyle \cos \left(\alpha - \beta\right) = \cos \alpha .\cos \beta + \sin \alpha .\sin \beta$

5. Originally Posted by Liam
I'm having trouble remembering the identites from the start of the course and am now thoroughly confused haha

If somebody could help me with these questions that would be great.

4. Solve for all possible values of X:
$\displaystyle -5sin(15(x-17degrees))-3=0$

excuse my lack of math typing skills
$\displaystyle - 5\sin \left[ {15\left( {x - 17^\circ } \right)} \right] - 3 = 0 \hfill \\$

$\displaystyle - 5\sin \left[ {15\left( {x - 17^\circ } \right)} \right] = 3 \hfill \\$

$\displaystyle \sin \left[ {15\left( {x - 17^\circ } \right)} \right] = \frac{3} {{ - 5}} \hfill \\$

$\displaystyle 15\left( {x - 17^\circ } \right) = \sin ^{ - 1} \left( { - \frac{3} {5}} \right) \hfill \\$

$\displaystyle {\text{Now, }}\sin ^{ - 1} \left( { - \frac{3} {5}} \right) = - 36.87^\circ = - 37^\circ \hfill \\$

$\displaystyle \sin \theta {\text{ is negative in 3rd and 4th quadrants, so, angle is}} \hfill \\$

$\displaystyle = 180^\circ + 37^\circ {\text{ and }} = 360^\circ - 37^\circ \hfill \\$

$\displaystyle = 217^\circ ,323^\circ \hfill \\$

$\displaystyle 15\left( {x - 17^\circ } \right) = 217^\circ \hfill \\$

$\displaystyle x - 17^\circ = 14.5^\circ \hfill \\$

$\displaystyle x = 14.5^\circ + 17^\circ = 31.5^\circ \hfill \\$

$\displaystyle {\text{Also,}} \hfill \\$

$\displaystyle 15\left( {x - 17^\circ } \right) = 323^\circ \hfill \\$

$\displaystyle {\text{Now, finish it, in the same way}}{\text{.}} \hfill \\$

6. Thanks Shyam, you were of great help!