Exact trigonometric relationships
Use the Pythagorean identity to find a missing squared ratio, then choose the square-root sign from the quadrant. Use the sine-over-cosine quotient only after the relevant sine and cosine values are known
\[\sin^2\theta+\cos^2\theta=1,\quad\tan\theta=\frac{\sin\theta}{\cos\theta}\]
- Keep both powers The Pythagorean identity applies to the squares of sine and cosine \(\displaystyle \sin^2\theta+\cos^2\theta=1\)
- Find the missing square Substitute the known ratio and rearrange before taking a square root \(\displaystyle \sin\theta=\frac{3}{5}\Rightarrow\cos^2\theta=1-\frac{9}{25}=\frac{16}{25}\)
- Use the quadrant sign Choose the sign of the square root from the quadrant containing the angle \(\displaystyle \theta\in\mathrm{QII}\Rightarrow\cos\theta=-\sqrt{\frac{16}{25}}=-\frac{4}{5}\)
- Use the quotient in order Tangent is sine divided by cosine \(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{3/5}{-4/5}=-\frac{3}{4}\)
Worked example
For the given cosine and quadrant, find sine and tangent exactly
\[\cos\theta=-\frac{5}{13},\quad\theta\in\mathrm{QIII}\]
- \(\displaystyle \sin^2\theta=1-\left(-\frac{5}{13}\right)^2=1-\frac{25}{169}=\frac{144}{169}\)Rearrange the Pythagorean identity and substitute the known cosine
- \(\displaystyle \sin\theta=-\sqrt{\frac{144}{169}}=-\frac{12}{13}\)Sine is negative in QIII, so choose the negative square root
- \(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-12/13}{-5/13}=\frac{12}{5}\)Divide sine by cosine; the two negative signs give a positive tangent
Common mistakes
Dropping the powers
✕\(\displaystyle \sin\theta+\cos\theta=1\)✓\(\displaystyle \sin^2\theta+\cos^2\theta=1\)
The Pythagorean identity applies to the squares of sine and cosine, not their unsquared sumIgnoring the quadrant sign
✕\(\displaystyle \theta\in\mathrm{QII},\quad\cos\theta=\sqrt{\frac{16}{25}}=\frac{4}{5}\)✓\(\displaystyle \theta\in\mathrm{QII},\quad\cos\theta=-\sqrt{\frac{16}{25}}=-\frac{4}{5}\)
Cosine is negative in QII, so the negative square-root branch is requiredInverting the tangent quotient
✕\(\displaystyle \tan\theta=\frac{\cos\theta}{\sin\theta}\)✓\(\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}\)
The quotient identity defines tangent as sine divided by cosineDifficulty levels
introUse the Pythagorean identity to find one missing exact ratio
standardDeduce related ratios using a given quadrant or complementary-angle relationship
challengeUse a quotient, verify an identity or combine several exact relationships
examApply identities, exact ratios and quadrant information in one problem
Questions about this topic
Why is a square root sign needed after using the identity?+
The identity gives a squared ratio, so both positive and negative roots are possible until the quadrant determines the sign
How is the sign of a trigonometric ratio chosen?+
Use the quadrant: sine is positive above the horizontal axis, cosine is positive to the right, and tangent has the sign of their quotient
What is the quotient identity for tangent?+
Tangent is sine divided by cosine, provided cosine is non-zero
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 10: Transformed circular functions