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:
$cos2A/(cosA+sinA)$

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

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

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

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

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

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

$=\cos A - \sin A$

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

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

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:
$sin^2(2)A=sin^2A-sin^4A$
Right Side $= \sin^2 A(1 - \sin^2 A)$

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

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

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

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

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

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. $sqrt(2)cos(A-(pi/4))=cosA+sinA$
$=\sqrt{2} \cos \left(A - \frac{\pi}{4}\right)$

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

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

$= \cos A + \sin A$

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

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:
$-5sin(15(x-17degrees))-3=0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

6. Thanks Shyam, you were of great help!