trigo>angle

**sanikui** · September 11th 2008, 08:07 PM

sinA+3cosA=2

Angle between 0* to 360*

Can use this method Rsin(A+a) to find , or??

THX FOR WHO HELPING

**Shyam** · September 11th 2008, 08:39 PM

Originally Posted by sanikui

sinA+3cosA=2

Angle between 0* to 360*

Can use this method Rsin(A+a) to find , or??

THX FOR WHO HELPING

$\sin A + 3\cos A = 2 \hfill \\$

$\Rightarrow 3\cos A = 2 - \sin A \hfill \\$

$\Rightarrow 3\sqrt {1 - \sin ^2 A} = 2 - \sin A{\text{ (because }}\cos^2 A = 1 - \sin ^2 A) \hfill \\$

${\text{squaring both sides,}} \hfill \\$

$\Rightarrow 9\left( {1 - \sin ^2 A} \right) = \left( {2 - \sin A} \right)^2 \hfill \\$

$\Rightarrow 9 - 9\sin ^2 A = 4 - 4\sin A + \sin ^2 A \hfill \\$

$\Rightarrow 10\sin ^2 A - 4\sin A - 5 = 0 \hfill \\$

${\text{This is a quadratic equation in sinA, solve by using quadratic formula, }}$

$\sin A = \frac{{ - b \pm \sqrt {b^2 - 4ac} }} {{2a}} \hfill \\$

$\sin A = \frac{{4 \pm \sqrt {\left( { - 4} \right)^2 - 4\left( {10} \right)\left( { - 5} \right)} }} {{2\left( {10} \right)}} \hfill \\$

$\sin A = \frac{{4 \pm \sqrt {216} }} {{20}} = \frac{{4 \pm 6\sqrt 6 }} {{20}} \hfill \\$

$\sin A = \frac{{2 \pm 3\sqrt 6 }} {{10}} = 0.934846922,{\text{ }} - 0.534846922 \hfill \\$

$\Rightarrow {\text{For positive value, }}A = 69^\circ ,{\text{ }}111^\circ \hfill \\$

$\Rightarrow {\text{For negative value, }}A = 212^\circ ,{\text{ }}328^\circ \hfill \\$

**sanikui** · September 11th 2008, 09:00 PM

why cosA not same as cosA= 1-2sin^2 (A/2)??

**Soroban** · September 11th 2008, 09:16 PM

Hello, sanikui!

$\sin A+3\cos A\:=\:2,\quad 0^o \leq A < 360^o$

Divide by $\sqrt{10}\!:\quad\frac{1}{\sqrt{10}}\sin A + \frac{3}{\sqrt{10}}\cos A \:=\:\frac{2}{\sqrt{10}}$ .[1]

Let $\theta$ be an angle in a right triangle with: . $opp = 3,\:adj = 1,\;hyp = \sqrt{10}$
. . Then: . $\sin\theta = \frac{3}{\sqrt{10}},\;\cos\theta = \frac{1}{\sqrt{10}}$

Then [1] becomes: . $\cos\theta\sin A + \sin\theta\cos A \:=\:\frac{2}{\sqrt{10}}$
. . and we have: . $\sin(A + \theta) \:=\:\frac{2}{\sqrt{10}}$

Then: . $A + \theta \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) \quad\Rightarrow\quad A \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) - \theta$

Hence: . $A \;=\;\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) - \sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$

. . $\sin^{-1}\left(\frac{2}{\sqrt{10}}\right) \;=\;39.23^o \quad\Rightarrow\quad\begin{Bmatrix}140.77^o \\ 399.23^o \end{Bmatrix}$

. . $\sin^{-1}\left(\frac{3}{\sqrt{10}}\right) \;=\;71.57^o$

Therefore: . $A \;=\;\begin{Bmatrix}140.77^o - 71.57^o &=& \boxed{69.2^o} \\ \\[-3mm] 399.23^o - 71.57^o &=& \boxed{327.66^o} \end{Bmatrix}$

**Shyam** · September 11th 2008, 09:41 PM

Originally Posted by sanikui

why cosA not same as cosA= 1-2sin^2 (A/2)??

If you use that formula, and split it in half-angle formula, then you have to split sinA also in half-angle formula. After doing this you still have both sin and cos in the equation and you cannot solve it.

because you have to make cosA in terms of sinA, so that you can make an equation which has only one variable, so that you can solve it.

so you have to use, $\cos^2 A = 1 - \sin ^2 A$

**sanikui** · September 11th 2008, 10:28 PM

Originally Posted by Shyam

If you use that formula, and split it in half-angle formula, then you have to split sinA also in half-angle formula. After doing this you still have both sin and cos in the equation and you cannot solve it.

because you have to make cosA in terms of sinA, so that you can make an equation which has only one variable, so that you can solve it.

so you have to use, $\cos^2 A = 1 - \sin ^2 A$

I understood. Thank you~!V^^

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