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Trigonometry · Guided Practice

Trigonometry

27 questions with answers and video solutions. Try each one before revealing the answer.

1
Find the exact value of cos⁡75∘\displaystyle \cos 75^\circ.
Video solution coming soon
2
Find the exact value of sin⁡195∘\displaystyle \sin 195^\circ.
Video solution coming soon
3
Given that sin⁡x=513, 0<x<π2\displaystyle \sin x=\frac{5}{13},\,0 < x < \frac{\pi}{2}, find the exact value of sin⁡2x\displaystyle \sin 2x.
Video solution coming soon
4
An angle x\displaystyle x is such that 0<x<π4\displaystyle 0 < x < \frac{\pi}{4} and tan⁡2x=43\displaystyle \tan 2x=\frac{4}{3}.
(a) Find the exact value of cos⁡2x\displaystyle \cos 2x.
(b) Hence find the exact value of cos⁡x\displaystyle \cos x.
Video solution coming soon
5
Given that cos⁡2x=1213, 0<x<π4\displaystyle \cos 2x=\frac{12}{13},\,0 < x < \frac{\pi}{4}, find the exact value of tan⁡x\displaystyle \tan x.
Video solution coming soon
6
Given that 0<x<45∘\displaystyle 0 < x < 45^\circ and sin⁡x=35\displaystyle \sin x=\frac{3}{5}, determine the exact values of:
(a) cos⁡x\displaystyle \cos x
(b) sin⁡2x\displaystyle \sin 2x
(c) cos⁡2x\displaystyle \cos 2x
(d) cos⁡3x\displaystyle \cos 3x
(e) sin⁡3x\displaystyle \sin 3x.
Video solution coming soon
7
Solve cos⁡2x=sin⁡x\displaystyle \cos 2x=\sin x for 0⩽x<2π\displaystyle 0\leqslant x < 2\pi.
Video solution coming soon
8
Solve 2cos⁡2x∘+1=0\displaystyle 2\cos 2x^\circ+1=0 for 0⩽x<360\displaystyle 0\leqslant x < 360.
Video solution coming soon
9
Solve 2cos⁡2x∘=1\displaystyle 2\cos^2 x^\circ=1 for 0⩽x<360\displaystyle 0\leqslant x < 360.
Video solution coming soon
10
2019 P1 Q15
(4, 1)5 Marks

(a)  Solve the equation
sin⁡2x∘+6cos⁡x∘=0\displaystyle \sin 2x^{\circ}+6\cos x^{\circ}=0
for 0≤x<360.\displaystyle 0\le x \lt 360.
(b)  Hence solve
sin⁡4x∘+6cos⁡2x∘=0\displaystyle \sin 4x^{\circ}+6\cos 2x^{\circ}=0
for 0≤x<360.\displaystyle 0\le x \lt 360.

11
2021 P2 Q8
Solve the equation 2sin⁡(3x−60)∘+1=0, 0⩽x<180\displaystyle 2\sin(3x-60)^\circ+1=0,\:0\leqslant x < 180.
Video solution coming soon
12
2023 P2 Q7
5 Marks
Solve the equation
sin⁡x∘+2=3cos⁡2x∘\displaystyle \sin x^{\circ}+2=3\cos 2x^{\circ}
for 0≤x<360.\displaystyle 0\le x \lt 360.
13
Express 3cos⁡x∘+sin⁡x∘\displaystyle \sqrt{3}\cos x^\circ+\sin x^\circ in the form kcos⁡(x−a)∘\displaystyle k\cos(x-a)^\circ where k>0\displaystyle k > 0 and 0<a<360\displaystyle 0 < a < 360.
Video solution coming soon
14
Express 2cos⁡x∘−3sin⁡x∘\displaystyle 2\cos x^\circ-3\sin x^\circ in the form kcos⁡(x−a)∘\displaystyle k\cos(x-a)^\circ where k>0\displaystyle k > 0 and 0<a<360\displaystyle 0 < a < 360.
Video solution coming soon
15
Express 5sin⁡x∘−2cos⁡x∘\displaystyle 5\sin x^\circ-2\cos x^\circ in the form kcos⁡(x+a)∘\displaystyle k\cos(x+a)^\circ where k>0\displaystyle k > 0 and 0<a<360\displaystyle 0 < a < 360.
Video solution coming soon
16
Express sin⁡x−3cos⁡x\displaystyle \sin x-\sqrt{3}\cos x in the form ksin⁡(x+a)\displaystyle k\sin(x+a) where k>0\displaystyle k > 0 and 0<a<2π\displaystyle 0 < a < 2\pi.
Video solution coming soon
17
Express −2sin⁡x+7cos⁡x\displaystyle -2\sin x+7\cos x in the form ksin⁡(x−a)\displaystyle k\sin(x-a) where k>0\displaystyle k > 0 and 0<a<2π\displaystyle 0 < a < 2\pi.
Video solution coming soon
18
(a) Express 3cos⁡x∘+sin⁡x∘\displaystyle 3\cos x^\circ+\sin x^\circ in the form kcos⁡(x−a)∘\displaystyle k\cos(x-a)^\circ where k>0\displaystyle k > 0 and 0<a<360\displaystyle 0 < a < 360.
(b) State the maximum value of 3cos⁡x∘+sin⁡x∘\displaystyle 3\cos x^\circ+\sin x^\circ and find the value of 0⩽x<360\displaystyle 0\leqslant x < 360 at which it occurs.
(c) State the minimum value of 3cos⁡x∘+sin⁡x∘\displaystyle 3\cos x^\circ+\sin x^\circ and find the value of 0⩽x<360\displaystyle 0\leqslant x < 360 at which it occurs.
Video solution coming soon
19
Solve −3cos⁡x∘+2sin⁡x∘=−1\displaystyle -3\cos x^\circ+2\sin x^\circ=-1, where 0⩽x<360\displaystyle 0\leqslant x < 360.
Video solution coming soon
20
Solve 5sin⁡2x−4cos⁡2x=3\displaystyle 5\sin 2x-4\cos 2x=3, where 0⩽x<2π\displaystyle 0\leqslant x < 2\pi.
Video solution coming soon
21
2018 Spec P2 Q11
Show that sin⁡2x2cos⁡x−sin⁡xcos⁡2x=sin⁡3x\displaystyle \frac{\sin 2x}{2\cos x} - \sin x\cos^2 x = \sin^3 x, where 0<x<π2\displaystyle 0 < x < \frac{\pi}{2}.
Video solution coming soon
22
2024 P2 Q12
5 Marks
Solve the equation
2sin⁡2x∘−sin⁡2x∘=0\displaystyle 2 \sin 2x^{\circ}-\sin^{2}x^{\circ} = 0
for 0≤x<360\displaystyle 0\le x \lt 360.
23
2025 P2 Q11
4 Marks
Solve
3 sin⁡ 2x∘+4 cos⁡ x∘=0\displaystyle 3~\sin~2x^{\circ}+4~\cos~x^{\circ}=0
for 0≤x<360\displaystyle 0\le x \lt 360.
24
2026 P1 Q4
(1, 4, 2, 1)8 Marks

The diagram shows two right-angled triangles with angles p and q as marked.

Two right-angled triangles, one with sides 3 and 4 and angle p, the other with sides 13 and angle q

(a)  Determine the exact value of sin⁡q.\displaystyle \sin q.
(b)  Find the exact value of:
    (i)  sin⁡(p+q)\displaystyle \sin(p+q)
    (ii)  cos⁡(p+q).\displaystyle \cos(p+q).
(c)  Hence find the exact value of tan⁡(p+q).\displaystyle \tan(p+q).

25
2026 P2 Q10
5 Marks
Solve the equation 3cos⁡2x∘+10cos⁡x∘=1\displaystyle 3\cos 2x^{\circ}+10\cos x^{\circ}=1 for 0≤x<360.\displaystyle 0\le x \lt 360.
26
2026 P2 Q13
(4, 3)7 Marks

(a)  Express 2sin⁡x∘−7cos⁡x∘\displaystyle 2\sin x^{\circ}-7\cos x^{\circ} in the form ksin⁡(x−a)∘\displaystyle k\sin(x-a)^{\circ}, where k>0\displaystyle k \gt 0 and 0<a<360.\displaystyle 0 \lt a \lt 360.
The diagram shows part of the curve with equation y=2sin⁡x∘−7cos⁡x∘\displaystyle y=2\sin x^{\circ}-7\cos x^{\circ} and the line with equation y=3.\displaystyle y=3.
The curve and the line intersect at the point P.

Curve y = 2 sin x - 7 cos x meeting the line y = 3 at the point P

(b)  Find the x-coordinate of P.

27
2026 P2 Q15
3 Marks
Express (3sin⁡θ+cos⁡θ)(sin⁡θ+3cos⁡θ)\displaystyle (3\sin\theta+\cos\theta)(\sin\theta+3\cos\theta) in the form a+bsin⁡cθ\displaystyle a+b\sin c\theta, where a, b and c are integers.

Practice questions courtesy of Maths.scot. Full written solutions are on his site — the links above go straight to them.

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