EXPLORING TRIGONOMETRY QUESTIONS AND ANSWERS GRADE 12
Activity 1
- ∆ MNP is a right-angled triangle. Write down the trig ratios for:
- sin α
- sin β (4)
- tan β
- cos α (3)
- If MP = 13 and NP = 5, calculate cos β.

[7]
Solutions- sin α = MN(1)
MP - sin β = NP(1)
MP - tan β = NP(1)
MN - cos α = NP(1)
MP(4)
- MP = 13 and NP = 5, so we can find MP,
MP2 = MN2 + NP2 ……….Pythagoras
132 = MN2 + 52
169 = MN2 + 25
MN2 = 169 – 25
MN2 = 144
∴MN = 12
cos β = MN = 12
MPÂ Â 13(3)
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Activity 2
- If sin θ is negative and cos θ is positive, then which statement is true?
- 0° < θ < 90°
- 90° < θ < 180°
- 180° < θ < 270°
- 270° < θ < 360° (1)
- If tan θ < 0 and cos θ < 0, then which statement is true?
- 0° < θ < 90°
- 90° < θ < 180°
- 180° < θ < 270°
- 270° < θ < 360° (1)
- Will the following trig ratios be positive or negative?
- sin 315°
- cos (–215°)
- tan 215°
- cos 390° (4)
[6]
Solutions- Sin θ is negative in 3rd and 4th quadrants; cos θ is positive in 1st and 4th quadrants.
So θ is in the 4th quadrant. D. 270° < θ < 360° 3 (1) - tan θ < 0 in 2nd and 4th quadrants; cos θ < 0 in 2nd and 3rd quadrants.
So θ is in the 2nd quadrant. B. 90° < θ < 180° 3 (1) - sin 315° is in 4th quadrant so it is negative. 3 (1)
- cos (–215°) is in 2nd quadrant so it is negative. 3 (1)
- tan 215° is in 3rd quadrant, so it is positive. 3 (1)
- cos 390° is the same as cos 30° in the 1st quadrant, so it is positive. 3 (1)
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Activity 3
If cos β = p/√5 where p < 0 and β ∈ [180°; 360°], determine, using a diagram, an expression in terms of p for:
- tan β
- 2 cos2β – 1
[6]
Solutions- cos β =p/√5 = x/r = ; so x = p and r = √5
By Pythagoras, y2 = r2 – x2
∴ y2 = (√5)2 – p2
= 5 – p2
∴ y = ± √5 – p2
∴ y = – √5 – p2 since β is in quadrant 3, y is negative
∴ tan β = –√5 – p2 (4)
p
 - 2 cos2β – 1 = 2 ( p/√5)2 – 1
2p 2 -1 (2) [6]
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Activity 4
Without using a calculator, determine the value of:
- cos 150°
- sin (–45°)
- tan 480°
[7]
Solutions- cos 150° rewrite as (180 – ?)
= cos(180° – 30°) quadrant II, cos θ negative
= –cos 30° special ratios
= –√3/2 (2) - sin(–45°) sin(–θ) = –sin θ; quadrant IV, sin θ negative
= –sin 45° special ratios
= – 1/√2 (2) - tan 480° write as an angle in the first rotation of 360°
= tan (480° – 360°)
= tan 120° quadrant II, rewrite as (180 – ?)
= tan (180° – 60°) tan θ negative
= –tan 60° special ratios
= –√3 (3)
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Activity 5
Write the trig ratios as the trig ratios of their co-functions:
- sin 50°
- cos 70°
- sin 100°
- cos 140°
[4]
Solutions- sin 50° = sin(90° – 40°) = cos 40°
- cos 70° = cos(90° – 20°) = sin 20°
- sin 100° = sin(90° + 10) = cos 10°
- cos 140° = cos(90° + 50°) = –sin 50°
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Summary
Any angle (obtuse or reflex) can be reduced to an acute angle by using:
- Convert negative angles to positive angles
- Reduce angles greater than 360°
- Use reduction formulae
- Use co-functions
Activity 6
Simplify without using a calculator:
- sin(180° + x). cos 330°.tan 150° (4)
sin x -    cos 750°.tan 315°.cos(–θ)   (8)
cos(360°- θ).sin300°.sin(180°- θ) - tan 480°.sin 300°.cos 14°.sin(–135°) (9)
sin104°.cos225° -   cos 260°.cos 170°  (7)
sin10°.sin190°.cos350°
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Activity 7
Simplify the following expressions.
-    cos (180°–x) sin (x –90°) – 1  (8)
tan2(540° + x) sin(90°+x)cos(–x) - [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)] (3)
- cos2θ (1 + tan2θ) (3)
- 1 – cos2θ
1– sin2θ (3)
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Solutions-   cos (180°–x) sin (x –90°) – 1  (8)
tan2(540° + x) sin(90°+x)cos(–x)  – use reduction formulae and co-functions
=     (–cos x)(–cos x) –1
tan2(540°- 360°+ x) cos x. cos x
– multiply out numerator and denominator reduction of angle > 360°
=    cos2x–1
tan2(180°+ x).cos2x
– use trig identity format for cos2x – 1 reduction formula
=  –(1 – cos2x)
tan2Â x . cos2x
– use trig identities for 1 – cos2x and for tan x
=  –sin2x
sin2x . cos2x
cos2x   1
= sin2x = -1
sin2x
– simplify
(8) - [sin(–θ) + cos(360° + θ)][cos(θ – 90°) + cos(180°+θ)] – reduce to angle < 90°
=[–sin θ + cos θ][cos (–(90° – θ))+ (–cos θ)] – simplify; use co-functions
=(–sin θ + cos θ)(sin θ – cos θ) – multiply out using FOIL
= –sin2 θ + sin θ cos θ + cos θ sin θ – cos2 θ
= –(sin2θ + cos2θ) + 2 sin θ cos θ – use trig identity
= –1 + 2 sin θ cos θ – use double angle identity
= –1 + sin2θ (3) - cos2θ (1 + tan2θ) – multiply out the bracket
=cos2θ + cos2θ.tan2θ – use trig identity for tan θ
= cos2θ + cos2θ . sin2θ – simplify
1    cos2θ.
= cos2θ + sin2θ 3 = 1 – use trig identity sin2θ + cos2θ = 1 (3) - 1 – cos2θ. – use trig identity sin2θ + cos2θ = 1
1 – sin2θ
= sin2θ – use trig identity for tan θ
cos2θ
= tan²θ (3)
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Activity 8
Prove the following identities:
sin x ∙tan x + cos x =  1
cos x (4)
(sin x + tan x) (  sin x  ) = sin x. tan x (7)
1 +Â cos x
1   =  cos x  + tan x (6)
cos x  1 + sin x
1  + tan x = tanx  (5)
tanx        sin2x
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Hints for solving trig identities:
- Choose either the lefthand side or the righthand side and simplify it to look like the other side.
- If both sides look difficult, you can try to simplify on both sides until you reach a point where both sides are the same.
- It is usually helpful to write tan θ as sinθ
cosθ . - Sometimes you need to simplify sinθ to tan θ.
cos θ - If you have sin2x or cos2x with +1 or –1, use the squares identities (sin2θ + cos2θ = 1).
- Find a common denominator when fractions are added or subtracted.
- Factorise if necessary – specify with examples i.e. common factor, DOPS, Trinomial, sum/diff of two cubes
Activity 9
- If cos 20° = p, determine the following ratios in terms of p:
- cos 380°
- sin 110°
- sin 200° (6)
- Determine the general solution for x in the following equations:
- 5 sin x = cos 320° (correct to 2 decimal places)
- 3 tan x + √3 = 0 (without using a calculator)
- tan x–1 = –3 (correct to one decimal place) (10)
2
- Determine x for x ∈[–180°; 180°] if 2 + cos (2x – 10°) = 2,537 (6)
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Solutions- cos 20° = p/1so x = p and r = 1
By Pythagoras, y2 = r2 – x2
y2 = 12 – p2 = 1– p2
y = √1 – p2 first quadrant, so y is positive- cos 380° = cos (360° + 20°) = cos 20° = p (2)
- sin 110° reduction formula
= sin (180° – 70°)
= sin 70° 3 co-function
= sin (90° – 20°)
= cos 20° 3 = p 3 (3) - sin 200° = sin (180° + 20°)
= –sin20°
= – √1 – p2 = –√1 – p2  (1) (6)
- 5 sin x = cos 320°
5 sin x = 0,766044
sin x = 0,15320…
Ref angle = 8,81°
x = 8,81° + k360° OR x = 180° – 8,81° + k360°
x = 171,19° + k360° k ∈ Z (4)
Calculator keys:
cos 320 =
÷ 5 =
SHIFT sin ANS = - 3 tan x + √3 = 0
3 tan x = – √3
tan x = – √3/3 [special angle: tan 30° tan 30° = – √3/3]
Ref angle = 30°
x = 180° – 30° + k180°
x = 150° + k180° 3 k ∈ Z (3) - tan x–1 = –3 multiply both sides by 2
2
tan x – 1 = –6
tan x = –5 reference angle is 78,69…°
∴x = 180° –78,69…° + k180°
x = 101,31° + k180°; k ∈ Z (3) (10)
- 2 + cos (2x – 10°) = 2,537
cos (2x – 10°) = 0,537
Ref angle = 57,52…..°
2x – 10° = 57,52…..° + k360° or 2x – 10° = 360° – 57,52° + k360°
[solve equations]
2x = 67,52….° + k360° or 2x = 312,48…° + k360°
[divide all terms on both sides by 2]
x = 33,76° + k180° or x = 156,24° + k180° 3 k ∈ Z
x ∈ [–180°; 180°]
So for k = –1: x = 33,76° –180° = –146,24° or x = 156,24° – 180° = –23,76°
For k = 0: x = 33,76° or x = 156,24°
(For k = 1, x will be > 180°, so it is too big)
Solution: x ∈ {–146,24°; –28,76°; 33,76°; 156,24°} (6)
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Activity 10
Do NOT use a calculator to answer this question. Show ALL calculations. Prove that:
- cos 75° =√2 ( √3 –1) (5)
4 - Prove that cos(90° – 2x).tan(180° + x) + sin2(360° – x) = 3sin2x (7)
- Prove that (tan x – 1)(sin 2x – 2cos2x) = 2(1 – 2sin x cos x) (7)
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- Solutions
1. LHS = cos 75° = cos(45° + 30°)
= cos45°.cos30° – sin45°.sin30°
= √2/2 . √3/√2 – √2/2 . ½
= √2 . √3 – √2/4
4
=√2 (√3 –1 = RHS (5)
4 - LHS = cos(90° – 2x).tan(180° + x) + sin2(360° – x) co-functions and reductions
= sin2 x. tan x + sin2 x                        double angle for sin 2x
trig identity for tan x
= 2sin x.cos x. sin x + sin2 x simplify
cos x
= 2 sin2 x + sin2 x
= 3 sin2x = RHS (7) - There are several ways to prove this. Here is one solution.
LHS = (tan x – 1)(sin 2x – 2cos2x)
= (sin x – 1 ) (2sin x. cos x – 2cos2x) double angle identity for sin 2x
cos x
= 2sin2 x – 2sin x. cos x – 2sin x. cos x + 2cos2x multiply out
= 2 sin2 x – 4 sin x cos x + 2 cos2 x
= 2(sin2 x – 2sin x. cos x + cos2x)       trig identity sin2 x + cos2 x = 1
= 2(1 – 2sin x. cos x) = RHS (7)
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Activity 11
Determine the general solution for x in the following:
- sin 2x. cos 10° – cos 2x. sin 10° = cos 3x (8)
- cos2 x = 3 sin 2x (11)
- 2sinx = sin(x + 30°) (5)
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Solutions- sin 2x. cos 10° – cos 2x. sin 10° = cos 3x use compound angle identity
∴ sin (2x – 10°) = cos 3x use co-functions
∴ sin (2x – 10°) = sin (90° – 3x)
∴ 2x – 10° = 90°–3x + k360°3or 2x–10° = 180°–(90°–3x) + k360°3 k ∈ Z
∴ 5x = 100° + k360° 2x – 10° = 90° + 3x + k360°
∴ x = 20° + k72° –x = 100 + k360°3
x = –100 – k360 3 k ∈ Z (8) - cos2 x = 3 sin 2x use double angles for sin 2x
cos2 x = 3(2 sin x.cos x)3 make LHS = 0
cos2 x – 3(2 sin x.cos x) = 0 multiply out
cos2 x – 6 sin x.cos x = 0 common factor
cos x (cos x – 6 sin x) = 0
∴ cos x = 03 or cos x – 6 sin x = 0
cos x = 0 or cos x = 6sin x
cos x   cos x
cos x = 0 or 1 = 6 tan x
cos x = 0 or tan x =Â 1/6
Reference angle = 90° or reference angle = 9,46°
∴ x = 90° + k360° or x = 360°–90° + k360° or x = 9,46° + k180° 3k ∈ Z
x = 270° + k360° or x = 180° + 9,46° + k360° k ∈ Z
= 189,46° + k360° k ∈ Z (11) - 2 sin x = sin ( x + 30° )
2 sin x = sin x. cos30° + cosx.sin30°3
2 sin x = sin x. √3/2  + cos x. ½ multiply by 2
4 sin x = √3 sin x + cos x divide by cos x
4 tan x = √3 tan x + 1
4 tan x − √3 tan x = 1
tan x =Â Â 1
4 − √3
x = 23,79° + k180°; k ∈ Z 3 (5)
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