smc-1025-10-Double Angles

Trigonometry, EXT1 T2 2024 SPEC2 4 MC

Given that \(\sin (x)=a\), where \(x \in\left(\dfrac{3 \pi}{2}, 2 \pi\right)\), then \(\cos \left(\dfrac{x}{2}\right)\) is equal to

\(-\dfrac{\sqrt{1+\sqrt{1-a^2}}}{\sqrt{2}}\)
\(\dfrac{\sqrt{1-\sqrt{a^2-1}}}{\sqrt{2}}\)
\(\dfrac{\sqrt{1+\sqrt{1-a^2}}}{\sqrt{2}}\)
\(-\dfrac{\sqrt{\sqrt{1-a^2}-1}}{\sqrt{2}}\)

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\(A\)

Show Worked Solution

\(\cos^2(x)+\sin^2(x)\)	\(=1\)
\(\cos^2(x)+a^2\)	\(=1\)
\(\cos(x)\)	\(=\sqrt{1-a^2}\ \ \Big(\text{take +ve root since}\ x \in\left(\frac{3 \pi}{2}, 2\pi\right),\ \cos(x)>0 \Big)\)
\(2\cos^2 \left(\dfrac{x}{2}\right)-1\)	\(=\sqrt{1-a^2}\)
\(\cos^2 \left(\dfrac{x}{2}\right)\)	\(=\dfrac{\sqrt{1-a^2}+1}{2}\)
\(\cos\left(\dfrac{x}{2}\right)\)	\(=\pm \sqrt{\dfrac{\sqrt{1-a^2}+1}{2}}\)

\(x \in\left(\dfrac{3 \pi}{2}, 2 \pi\right)\ \ \Rightarrow \ \ \dfrac{x}{2} \in\left(\dfrac{3 \pi}{4}, \pi\right) \)

\(\cos \left( \dfrac{x}{2} \right) \lt 0\ \ \text{(take negative root)}\)

\(\Rightarrow A\)

♦♦♦ Mean mark 27%.

Trigonometry, EXT1 T2 EQ-Bank 8

Show that \(\dfrac{\sin 2 x}{1+\cos 2 x}=\tan x\). (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Hence find the exact value of \(\tan \dfrac{\pi}{8}\) in simplest form. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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a. \(\dfrac{\sin 2 x}{1+\cos 2 x}\)	\(=\dfrac{2\sin x \cos x}{2\cos^{2} x}\)
	\(=\dfrac{\sin x}{\cos x}\)
	\(=\tan x\)

b. \(\sqrt2-1\)

Show Worked Solution

a. \(\dfrac{\sin 2 x}{1+\cos 2 x}\)	\(=\dfrac{2\sin x\, \cos x}{2\cos^{2} x}\)
	\(=\dfrac{\sin x}{\cos x}\)
	\(=\tan x\)

b.	\(\tan \dfrac{\pi}{8}\)	\(=\dfrac{\sin \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}\)
		\(= \dfrac{\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}} \times \dfrac{\sqrt2}{\sqrt2} \)
		\(= \dfrac{1}{\sqrt2+1} \times \dfrac{\sqrt2-1}{\sqrt2-1}\)
		\(=\sqrt2-1\)

Trigonometry, EXT1 T2 2022 SPEC2 2 MC

The expression `1-\frac{4\sin^2(x)}{\tan^2(x)+1}` simplifies to

`1-2\cos^2(2x)`
`2\sin(2x)`
`2\sin^2(2x)`
`\cos^2(2x)`

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`D`

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`1-\frac{4 \sin ^2 x}{\tan ^2 x+1}`	`= 1-\frac{4 \sin ^2 x}{\sec ^2 x}`
	`= 1-(2 \sin x\ \cos x)^2`
	`= 1-\sin ^2 (2 x)`
	`= \cos ^2 (2 x)`

`=>D`

Trigonometry, EXT1 T2 2019 HSC 6 MC

It is given that `sin x = 1/4`, where `pi/2 < x < pi`.

What is the value of `sin 2x`?

A. `-7/8`

B. `-sqrt 15/8`

C. `sqrt 15/8`

D. `7/8`

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`B`

Show Worked Solution

`sin x = 1/4`

`cos x = -sqrt 15/4, \ \ (pi/2 < x < pi)`

`sin 2x`	`= 2 sin x cos x`
	`= 2 xx 1/4 xx – sqrt 15/4`
	`= – sqrt 15/8`

`=> B`

Trigonometry, EXT1 T2 2013 HSC 8 MC

The angle `theta` satisfies `sin theta = 5/13` and `pi/2 < theta < pi`.

What is the value of `sin 2 theta`?

`10/13`
`- 10/13`
`120/169`
`- 120/169`

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`D`

Show Worked Solution

♦ Mean mark 44%

`sin theta = 5/13\ \ \ (pi/2 < theta < pi)`

`text(S)text(ince)\ \ theta\ \ text(is in the 2nd quadrant,)`

`cos theta = -12/13`

`:.sin 2theta`	`= 2 sin theta cos theta`
	`= 2 xx 5/13 xx -12/13`
	`= -120/169`

`=> D`