Trignometric Identity Questions

**Liam** · November 8th 2008, 04:33 PM

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

**Shyam** · November 8th 2008, 04:46 PM

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

**Shyam** · November 8th 2008, 05:02 PM

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, $2 \sin A \cos A = \sin 2A$

**Shyam** · November 8th 2008, 05:14 PM

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

**Shyam** · November 8th 2008, 05:56 PM

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

$- 5\sin \left[ {15\left( {x - 17^\circ } \right)} \right] - 3 = 0 \hfill \\$

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

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

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

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

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

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

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

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

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

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

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

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

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

**Liam** · November 9th 2008, 03:42 PM

Thanks Shyam, you were of great help!

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