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Compound and multiple angles

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Compound and multiple angles - test bite

Question

If $\sin A\; = \;\frac{4}{5}$ and $\sin B\, = \;\frac{5} {{13}}$ where $0 < \;A < \frac{\pi } {2}$ and $0 < \;B < \frac{\pi } {2}$ , find the exact value of:

1) $\sin \left( {A + \;B} \right)$

2) $\cos \;\left( {A\; - \;B} \right)$

Answer

Sketch two right angled triangles and use Pythagoras' Theorem to calculate the third side.

Right angled triangle with sides 3, 4, 5

Right angled triangle with sides 12, 5, 13

$\sin \left( {A + \;B} \right)\; = \;\sin \;A\;\cos \;B\; + \;\cos \;A\;\sin \;B$

You must expand this.

Obtain the values from the triangles above.

$= \frac{4} {5} \times \frac{{12}} {{13}} + \frac{3} {5} \times \frac{5} {{13}}$

$= \frac{{48}} {{65}}\; + \;\frac{{15}} {{65}}$

$= \;\frac{{63}} {{65}}$

2 $\cos \left( {A - \;B} \right)\; = \;\cos A\;\cos \;B\; + \;\sin A\;\sin B$

Again, you must expand this.

Obtain the values from the triangles above.

$= \frac{3} {5} \times \frac{{12}} {{13}} + \frac{4} {5} \times \frac{5} {{13}}$

$= \frac{{36}} {{65}}\; + \;\frac{{20}} {{65}}$

$= \;\frac{{56}} {{65}}$

Question

Expand and simplify:

1) $\sin \left( {2x\; - \;90} \right)^ \circ$

2) $\cos \left( {3x\; + \;90} \right)^ \circ$

3) $\sin \left( {2\pi \; + \;x} \right)$

4) $\cos \left( {4x\; - \;2\pi } \right)$

Answer

To expand and simplify:

Use $\sin 90^ \circ = 1$ and $\cos 90^ \circ = 0$

$\sin 2\pi = 0$ and $\cos 2\pi = 1$

$\eqalign{ & \sin \left( {2x\; - \;90} \right)^ \circ = \;\sin 2x^ \circ \cos 90^ \circ \; - \;\cos 2x^ \circ \sin 90^ \circ \cr & = \; - \cos 2x^ \circ \cr}$

$\eqalign{ & \cos \left( {3x\; + \;90} \right)^ \circ = \;\cos 3x^ \circ \cos 90^ \circ \; - \;\sin 3x^ \circ \sin 90^ \circ \cr & = \; - \sin 3x^ \circ \cr}$

$\sin \left( {2\pi \; + \;x} \right) &=& \;\sin 2\pi \cos x\; + \;\cos 2\pi \sin x \hfill \\ &=& \;\sin x \hfill \\$

$\cos \left( {4x\; - \;2\pi } \right) &=& \;\cos 4x\cos 2\pi \; + \;\sin 4x\sin 2\pi \hfill \\ &=& \;\cos 4x \hfill \\$

Question

If $\cos x^ \circ \; = \frac{1} {{\sqrt 5 }}$ where $0 \leq \;x\; \leq \;90$ , find the exact value of :

1) $\cos \left( {2x} \right)^ \circ$

2) $\sin \left( {2x} \right)^ \circ$

Answer

Right angled triangle with sides 1, 2, square root 5

Make a sketch and use Pythagoras' Theorem to find the third side. Then expand using a double angle formula and substitute values from this triangle.

1) You can use any one of the three formulae for cos(2x)° here.

$\cos \left( {2x} \right)^ \circ = \cos ^2 x ^\circ - \sin ^2 x ^\circ$

$&=& \;\left( {\frac{1} {{\sqrt 5 }}} \right)^2 \; - \;\left( {\frac{2} {{\sqrt 5 }}} \right)^2 \hfill \\ &=& \;\frac{1} {5}\; - \;\frac{4} {5} \hfill \\ &=& \; - \;\frac{3} {5} \hfill \\$

$\sin \left( {2x} \right)^ \circ \; &=& \;2\sin x ^\circ \cos x ^\circ \hfill \\ &=& \;2 \times \frac{2} {{\sqrt 5 }} \times \frac{1} {{\sqrt 5 }}\; \hfill \\ &=& \;\frac{4} {5} \hfill \\$

Question: Given that $\sin \;(x)^ \circ = \;\frac{1} {{\sqrt 2 }}$ where , find the exact value of $\sin \left( {x\; - \;30} \right)^ \circ$

Answer

Right angled triangle with sides 1, 1, square root 5

From the triangle

$\sin x\; = \;\frac{1} {{\sqrt 2 }}$ and $\cos x\; = \;\frac{1} {{\sqrt 2 }}$ and so expand and use $\sin 30^ \circ = \frac{1} {2}$ and $\cos 30^ \circ = \frac{{\sqrt 3 }} {2}$

$\sin \;\left( {x\; - \;30} \right)^ \circ \; = \;\sin x^ \circ \cos 30^ \circ \; - \;\cos x^ \circ \sin 30^ \circ$

$&=& \frac{1} {{\sqrt 2 }} \times \frac{{\sqrt 3 }} {2}\;\;{\text{ - }}\;\frac{1} {{\sqrt 2 }} \times \frac{1} {2} \hfill \\ &=& \frac{{\sqrt 3 }} {{2\sqrt 2 }}\;\;{\text{ - }}\;\frac{1} {{2\sqrt 2 }} \hfill \\ &=& \;\frac{{\sqrt 3 \; - \;1}} {{2\sqrt 2 }} \hfill \\$

Question: Solve $\sin (2x)^ \circ \; = \;\cos \left( x \right)^ \circ$ for $0 < \;x\; < 360$

Answer

Since the equation involves sin 2x and cos x you should:

take all the terms to the left hand side
expand sin 2x
factorise and solve each factor

$\eqalign{ \sin 2x^\circ &=& \cos x^\circ \cr \sin 2x^\circ - \cos x^\circ &=& 0 \cr 2\sin x^\circ \cos x^\circ - \cos x^\circ &=& 0 \cr \cos x^\circ (2\sin x^\circ - 1) &=& 0 \cr}$

either cos x° = 0 or sin x° = 0.5

and x = 90 or 270 or x = 30 or 150

Solutions are {30, 90,150,270}

Question: Solve $\cos (2x) = \cos (x)$ for $0 \leq x \leq 2\pi$

Answer

Since the equation involves cos(2x) and cos(x)

take all the terms to the left hand side
replace cos(2x) with 2 cos²x - 1
factorise and solve each factor

$\eqalign{ \cos 2x &=& \cos x \cr \cos 2x - \cos x &=& 0 \cr 2\cos ^2 x - \cos x - 1 &=& 0 \cr (2\cos x + 1)(\cos x - 1) &=& 0 \cr}$ so either $\cos x = - \frac{1} {2}$ or $\cos x = 1$ and $x = \frac{{2\pi }} {3}$ or $\frac{{4\pi }} {3}$ or or $2\pi$

Solutions are {0, $\frac{{2\pi }} {3}$ , $\frac{{4\pi }} {3}$ , $2\pi$ }

Question: Solve $\cos (2x) - 3\sin (x) - 1 = 0$ for $0 \le x < 2\pi$

Answer

Since the equation involves cos2x and sin x

replace cos2x with $1 - 2\sin ^2 x$
factorise and solve each factor

$\eqalign{ \cos 2x - 3\sin x - 1 &=& 0\cr\cr 1 - 2\sin ^2 x - 3\sin x - 1 &=& 0\cr\cr 2\sin ^2 x + 3\sin x &=& 0\cr\cr \sin x(2\sin x + 3) &=& 0\cr}$ so either $\sin x = 0$ or $\sin x = - \frac{3} {2}$ there are no solutions for $\sin x = - \frac{3} {2}$ , since $- 1 \leq \sin x \leq 1$

Thus x = 0 or $\pi$ or $2\pi$

reject $2\pi$ since it is outside the domain

The solutions are {0 , $\pi$ }

Question: Solve $3\cos ^2 (x)^\circ - \cos (x)^\circ - 1 = 0$ for $0 \leq x \leq 360$

Answer

$3\cos ^2 (x)^\circ - \cos (x)^\circ - 1 = 0$ cannot be factorised so we need to use

quadratic formula $x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }} {{2a}}$ where a = 3, b = -1, c = -1 and $x \equiv \cos x$

$\eqalign{ \cos x^\circ &=& {{1 \pm \sqrt {1 + 12} } \over 6} \cr &=& {{1 \pm \sqrt {13} } \over 6} \cr}$

We need to solve

$\cos x^\circ = \frac{{1 + \sqrt {13} }} {6} \simeq 0.768$	or	$\cos x = \frac{{1 - \sqrt {13} }} {6} \simeq - 0.434$
so x = 39.8 or 320.2	or	x= 115.7 or 244.3

Question

Show that

$\cos (A - B) - \cos (A + B) = 2\sin A\sin B$

Answer: $LHS &=& \cos (A - B) - \cos (A + B) \hfill \\ &=& \cos A\cos B + \sin A\sin B - (\cos A\cos B - \sin A\sin B) \hfill \\ &=& \cos A\cos B + \sin A\sin B - \cos A\cos B + \sin A\sin B \hfill \\ &=& 2\sin A\sin B \hfill \\ &=& RHS \hfill \\$

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Compound and multiple angles - test bite

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