Ratios and Graphs of Trigonometric Functions · 6.1.4
Solving problems with trig ratios
Students combine what they know about trigonometric ratios, reference angles, quadrant signs and identities such as sin²θ + cos²θ = 1 to solve multi-step problems. This may involve finding an unknown ratio from given conditions, checking consistency between ratios, or working through a real or geometric scenario.
The official learning standard (6.1.4)
“Solve problems involving sine, cosine and tangent.”
What it means
Students combine what they know about trigonometric ratios, reference angles, quadrant signs and identities such as sin²θ + cos²θ = 1 to solve multi-step problems. This may involve finding an unknown ratio from given conditions, checking consistency between ratios, or working through a real or geometric scenario.
How it is examined
This standard appears mainly in Paper 2, where a question gives partial information (such as one ratio and the sign of another) and asks students to find a related ratio or angle, often requiring the Pythagorean identity. Paper 1 may test simpler combined-ratio calculations as short objective items.
Worked example
Given that sin x° = 0.8 and cos x° < 0, for 0° ≤ x ≤ 360°, find the value of tan x°.
- Since sin x is positive and cos x is negative, x lies in Quadrant II.
- Use the identity sin²x + cos²x = 1: cos²x = 1 - 0.8² = 1 - 0.64 = 0.36.
- cos x = -√0.36 = -0.6 (negative, since x is in Quadrant II).
- tan x = sin x ÷ cos x = 0.8 ÷ (-0.6) = -4/3.
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
Do I need to find the angle x itself, or just the ratio?
Read the question carefully, some ask only for a related ratio (like tan x here), which you can find directly using identities without ever calculating x. Others explicitly ask for the angle, in which case you still need the reference angle and quadrant rule afterwards.
Why is cos x negative here instead of positive?
The question states cos x° < 0, and Quadrant II is the only quadrant where sine is positive and cosine is negative at the same time. Taking the square root of cos²x gives ±0.6, and the sign condition tells you to keep the negative value, -0.6.
Can I use a calculator to find sin⁻¹(0.8) directly instead?
You could find the reference angle sin⁻¹(0.8) ≈ 53.13° and then work out x in Quadrant II, but that adds an unnecessary step here since the question only asks for tan x°. Using the Pythagorean identity directly is faster and avoids rounding errors from the angle.