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Practice set 1

Trigonometric Identities Practice

10 geometry practice problems covering sum and difference formulas, double-angle formulas, and auxiliary angle techniques.

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Use this set after the main trigonometry lesson if you want practice manipulating identities before moving on to graph analysis.

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1
Tangent Sum Formula Application

Problem

Given that tan⁡α=2\tan\alpha = 2 and 2sin⁡α=cos⁡(α−β)sin⁡β2\sin\alpha = \cos(\alpha-\beta)\sin\beta, find tan⁡β\tan\beta.

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Answer: 17\frac{1}{7}

Solution

Since sin⁡α=sin⁡[(α−β)+β]\sin\alpha = \sin[(\alpha-\beta)+\beta]:

sin⁡α=sin⁡(α−β)cos⁡β+cos⁡(α−β)sin⁡β\sin\alpha = \sin(\alpha-\beta)\cos\beta + \cos(\alpha-\beta)\sin\beta

From 2sin⁡α=cos⁡(α−β)sin⁡β2\sin\alpha = \cos(\alpha-\beta)\sin\beta:

2sin⁡(α−β)cos⁡β+2cos⁡(α−β)sin⁡β=cos⁡(α−β)sin⁡β2\sin(\alpha-\beta)\cos\beta + 2\cos(\alpha-\beta)\sin\beta = \cos(\alpha-\beta)\sin\beta

Therefore:

2sin⁡(α−β)cos⁡β=−cos⁡(α−β)sin⁡β2\sin(\alpha-\beta)\cos\beta = -\cos(\alpha-\beta)\sin\beta2tan⁡(α−β)=−tan⁡β2\tan(\alpha-\beta) = -\tan\beta

Since tan⁡α=2\tan\alpha = 2, we have tan⁡(α−β)=32tan⁡β\tan(\alpha-\beta) = \frac{3}{2}\tan\beta.

Using the tangent difference formula:

tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡α⋅tan⁡β=2−tan⁡β1+2tan⁡β\tan(\alpha-\beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta} = \frac{2 - \tan\beta}{1 + 2\tan\beta}

Solving: 2−tan⁡β=3+6tan⁡β2 - \tan\beta = 3 + 6\tan\beta, we get tan⁡β=17\tan\beta = \frac{1}{7}.

2
Cosine Sum Formula

Problem

Evaluate cos⁡147°cos⁡333°+cos⁡57°cos⁡63°\cos147°\cos333° + \cos57°\cos63°.

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Answer: −12-\frac{1}{2}

Solution

Using angle transformations:

cos⁡147°cos⁡333°+cos⁡57°cos⁡63°\cos147°\cos333° + \cos57°\cos63°=cos⁡(180°−33°)cos⁡(360°−27°)+cos⁡(90°−33°)cos⁡(90°−27°)= \cos(180°-33°)\cos(360°-27°) + \cos(90°-33°)\cos(90°-27°)=−cos⁡33°cos⁡27°+sin⁡33°sin⁡27°= -\cos33°\cos27° + \sin33°\sin27°=−(cos⁡33°cos⁡27°−sin⁡33°sin⁡27°)= -(\cos33°\cos27° - \sin33°\sin27°)=−cos⁡(33°+27°)=−cos⁡60°=−12= -\cos(33°+27°) = -\cos60° = -\frac{1}{2}
3
Roots and Tangent Formula

Problem

Given that tan⁡α\tan\alpha and tan⁡β\tan\beta are roots of x2−7x+13=0x^2 - 7x + 13 = 0, find tan⁡(α+β)\tan(\alpha+\beta).

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Answer: −712-\frac{7}{12}

Solution

By Vieta's formulas:

tan⁡α+tan⁡β=7,tan⁡α⋅tan⁡β=13\tan\alpha + \tan\beta = 7, \quad \tan\alpha\cdot\tan\beta = 13

Using the tangent sum formula:

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡α⋅tan⁡β=71−13=−712\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\cdot\tan\beta} = \frac{7}{1-13} = -\frac{7}{12}
4
Sine Sum Formula Application

Problem

Given sin⁡(2α−β)=513\sin(2\alpha-\beta) = \frac{5}{13} and cos⁡(α−β)=13sin⁡α\cos(\alpha-\beta) = \frac{1}{3}\sin\alpha, find sin⁡α\sin\alpha.

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Answer: 14\frac{1}{4}

Solution

We can write:

sin⁡(2α−β)=sin⁡[(α−β)+α]\sin(2\alpha-\beta) = \sin[(\alpha-\beta)+\alpha]=sin⁡(α−β)cos⁡α+cos⁡(α−β)sin⁡α= \sin(\alpha-\beta)\cos\alpha + \cos(\alpha-\beta)\sin\alpha=sin⁡(α−β)cos⁡α+13sin⁡2α= \sin(\alpha-\beta)\cos\alpha + \frac{1}{3}\sin^2\alpha

Therefore:

sin⁡(α−β)cos⁡α=513−13sin⁡α=112\sin(\alpha-\beta)\cos\alpha = \frac{5}{13} - \frac{1}{3}\sin\alpha = \frac{1}{12}

Thus:

sin⁡α=13⋅112=14\sin\alpha = \frac{1}{3}\cdot\frac{1}{12} = \frac{1}{4}
5
Sum of Squares Method

Problem

Given cos⁡α+cos⁡β=1010\cos\alpha + \cos\beta = \frac{\sqrt{10}}{10} and sin⁡α+sin⁡β=31010\sin\alpha + \sin\beta = \frac{3\sqrt{10}}{10}, find cos⁡(α−β)\cos(\alpha-\beta).

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Answer: −12-\frac{1}{2}

Solution

Squaring both equations:

cos⁡2α+2cos⁡αcos⁡β+cos⁡2β=110(1)\cos^2\alpha + 2\cos\alpha\cos\beta + \cos^2\beta = \frac{1}{10} \quad (1)sin⁡2α+2sin⁡αsin⁡β+sin⁡2β=910(2)\sin^2\alpha + 2\sin\alpha\sin\beta + \sin^2\beta = \frac{9}{10} \quad (2)

Adding (1) and (2):

1+2cos⁡αcos⁡β+2sin⁡αsin⁡β+1=11 + 2\cos\alpha\cos\beta + 2\sin\alpha\sin\beta + 1 = 12cos⁡(α−β)=−12\cos(\alpha-\beta) = -1cos⁡(α−β)=−12\cos(\alpha-\beta) = -\frac{1}{2}
6
Double-Angle Formula

Problem

Given sin⁡α−cos⁡α=13\sin\alpha - \cos\alpha = \frac{1}{3} and α∈(0,π)\alpha \in (0,\pi), find cos⁡2α\cos 2\alpha.

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Answer: −179-\frac{\sqrt{17}}{9}

Solution

Squaring sin⁡α−cos⁡α=13\sin\alpha - \cos\alpha = \frac{1}{3}:

1−2sin⁡αcos⁡α=191 - 2\sin\alpha\cos\alpha = \frac{1}{9}sin⁡2α=2sin⁡αcos⁡α=89>0\sin 2\alpha = 2\sin\alpha\cos\alpha = \frac{8}{9} > 0

Since α∈(0,π)\alpha \in (0,\pi) and sin⁡α,cos⁡α\sin\alpha, \cos\alpha have different signs, α∈(π2,π)\alpha \in (\frac{\pi}{2},\pi).

Therefore cos⁡α−sin⁡α<0\cos\alpha - \sin\alpha < 0, so:

cos⁡2α=(cos⁡α−sin⁡α)(cos⁡α+sin⁡α)<0\cos 2\alpha = (\cos\alpha - \sin\alpha)(\cos\alpha + \sin\alpha) < 0cos⁡2α=−1−sin⁡22α=−1−6481=−179\cos 2\alpha = -\sqrt{1 - \sin^2 2\alpha} = -\sqrt{1 - \frac{64}{81}} = -\frac{\sqrt{17}}{9}
7
Reduction and Double-Angle

Problem

Given sin⁡(α−π6)=14\sin\left(\alpha - \frac{\pi}{6}\right) = \frac{1}{4}, find sin⁡(2α+5π6)\sin\left(2\alpha + \frac{5\pi}{6}\right).

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Answer: 78\frac{7}{8}

Solution

Transform the angle:

sin⁡(2α+5π6)=sin⁡[2(α−π6)+π2+π]\sin\left(2\alpha + \frac{5\pi}{6}\right) = \sin\left[2\left(\alpha - \frac{\pi}{6}\right) + \frac{\pi}{2} + \pi\right]=−sin⁡[2(α−π6)+π2]= -\sin\left[2\left(\alpha - \frac{\pi}{6}\right) + \frac{\pi}{2}\right]=−cos⁡[2(α−π6)]= -\cos\left[2\left(\alpha - \frac{\pi}{6}\right)\right]=−(1−2sin⁡2(α−π6))= -\left(1 - 2\sin^2\left(\alpha - \frac{\pi}{6}\right)\right)=−1+2⋅(14)2=−1+18=78= -1 + 2 \cdot \left(\frac{1}{4}\right)^2 = -1 + \frac{1}{8} = \frac{7}{8}
8
Angle Evaluation

Problem

Evaluate 1+3tan⁡10°1+cos⁡20°\frac{1 + \sqrt{3}\tan 10°}{1 + \cos 20°}.

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Answer: 222\sqrt{2}

Solution

Rewrite the expression:

1+3tan⁡10°1+cos⁡20°=sin⁡10°+3cos⁡10°cos⁡10°⋅2cos⁡210°\frac{1 + \sqrt{3}\tan 10°}{1 + \cos 20°} = \frac{\sin 10° + \sqrt{3}\cos 10°}{\cos 10° \cdot 2\cos^2 10°}=2cos⁡(10°−60°)2sin⁡10°cos⁡10°⋅cos⁡10°= \frac{2\cos(10° - 60°)}{2\sin 10°\cos 10° \cdot \cos 10°}=2cos⁡(−50°)sin⁡20°⋅cos⁡10°= \frac{2\cos(-50°)}{\sin 20° \cdot \cos 10°}=2sin⁡40°sin⁡20°⋅cos⁡10°= \frac{2\sin 40°}{\sin 20° \cdot \cos 10°}=4sin⁡20°cos⁡20°sin⁡20°⋅cos⁡10°=4cos⁡20°cos⁡10°=22= \frac{4\sin 20°\cos 20°}{\sin 20° \cdot \cos 10°} = \frac{4\cos 20°}{\cos 10°} = 2\sqrt{2}
9
Auxiliary Angle Method

Problem

Given 2sin⁡α+1=23cos⁡α2\sin\alpha + 1 = 2\sqrt{3}\cos\alpha, find sin⁡(2α+π6)\sin\left(2\alpha + \frac{\pi}{6}\right).

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Answer: 78\frac{7}{8}

Solution

From 2sin⁡α+1=23cos⁡α2\sin\alpha + 1 = 2\sqrt{3}\cos\alpha:

4(sin⁡α+32cos⁡α)=14\left(\sin\alpha + \frac{\sqrt{3}}{2}\cos\alpha\right) = 1sin⁡(α+π3)=14\sin\left(\alpha + \frac{\pi}{3}\right) = \frac{1}{4}

Therefore:

sin⁡(2α+π6)=sin⁡[2(α+π3)−π2]\sin\left(2\alpha + \frac{\pi}{6}\right) = \sin\left[2\left(\alpha + \frac{\pi}{3}\right) - \frac{\pi}{2}\right]=−cos⁡[2(α+π3)]= -\cos\left[2\left(\alpha + \frac{\pi}{3}\right)\right]=−(1−2sin⁡2(α+π3))= -\left(1 - 2\sin^2\left(\alpha + \frac{\pi}{3}\right)\right)=2⋅(14)2−1=18−1=78= 2 \cdot \left(\frac{1}{4}\right)^2 - 1 = \frac{1}{8} - 1 = \frac{7}{8}
10
Value Finding Comprehensive

Problem

Given sin⁡α+sin⁡(α+π3)=33\sin\alpha + \sin\left(\alpha + \frac{\pi}{3}\right) = \frac{\sqrt{3}}{3}, find cos⁡(2α+π3)\cos\left(2\alpha + \frac{\pi}{3}\right).

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Answer: 79\frac{7}{9}

Solution

Expanding:

sin⁡α+sin⁡αcos⁡π3+cos⁡αsin⁡π3=33\sin\alpha + \sin\alpha\cos\frac{\pi}{3} + \cos\alpha\sin\frac{\pi}{3} = \frac{\sqrt{3}}{3}32sin⁡α+32cos⁡α=33\frac{3}{2}\sin\alpha + \frac{\sqrt{3}}{2}\cos\alpha = \frac{\sqrt{3}}{3}3sin⁡(α+π6)=33\sqrt{3}\sin\left(\alpha + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}sin⁡(α+π6)=13\sin\left(\alpha + \frac{\pi}{6}\right) = \frac{1}{3}

Therefore:

cos⁡(2α+π3)=1−2sin⁡2(α+π6)=1−29=79\cos\left(2\alpha + \frac{\pi}{3}\right) = 1 - 2\sin^2\left(\alpha + \frac{\pi}{6}\right) = 1 - \frac{2}{9} = \frac{7}{9}

Continue the practice path

Move from identity manipulation into graph reasoning

This set opens the trigonometry practice sequence. Review the lesson if needed, use the calculator for quick checks, or continue into graph-based problem solving.

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