In a right-angled triangle, \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}\).
By Pythagoras’ theorem, \((\text{opposite})^2 + (\text{adjacent})^2 = (\text{hypotenuse})^2\). Dividing through by \((\text{hypotenuse})^2\) gives \(\sin^2\theta + \cos^2\theta = 1\).
For tangent: \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\). Substituting sine and cosine gives \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).
Recipe (how to use it)
If you know \(\sin\theta\), you can find \(\cos\theta\) using \(\sin^2\theta + \cos^2\theta = 1\).
If you know \(\cos\theta\), you can find \(\sin\theta\).
If you know both sine and cosine, you can calculate tangent.
Spotting it
Look for questions where you are given one trigonometric ratio and asked to find another. If “prove” or “show that” appears, this identity is often the key.
Forgetting that cosine can be negative depending on the quadrant.
Using \(\tan = \sin \times \cos\) (wrong!) instead of \(\tan = \sin/\cos\).
Exam strategy
Write down the identity first: \(\sin^2\theta + \cos^2\theta = 1\).
Substitute the known value, then solve step by step.
For tangent, always divide sine by cosine.
Summary
The two most important trig identities at GCSE and A-level are \(\sin^2\theta + \cos^2\theta = 1\) and \(\tan\theta = \sin\theta/\cos\theta\). They connect all three trig functions and are essential for solving trig equations, proving identities, and working with right-angled triangles.
Worked examples
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\( If sin θ = 3/5, find cos θ. \)
\( sin²θ+cos²θ=1 \)
\( (3/5)²+cos²θ=1 \)
\( 9/25+cos²θ=1 \)
\( cos²θ=16/25 \)
\( cosθ=±4/5 \)
Answer:
±4/5
\( If cos θ = 12/13, find sin θ. \)
\( sin²θ+cos²θ=1 \)
\( sin²θ+(12/13)²=1 \)
\( sin²θ=25/169 \)
\( sinθ=±5/13 \)
Answer:
±5/13
\( sin θ = 0.6, cos θ = 0.8. Find tan θ. \)
\( tanθ=sinθ/cosθ=0.6/0.8=0.75 \)
Answer:
0.75
\( Prove that tan²θ+1=sec²θ using sin²+cos²=1. \)
\( tan²θ=(sin²θ)/(cos²θ) \)
\( tan²θ+1=(sin²θ+cos²θ)/(cos²θ)=1/cos²θ=sec²θ \)
Answer:
Identity proved
\( If sin θ = 4/5, find tan θ. \)
\( cos²θ=1-(4/5)²=9/25 \)
\( cosθ=±3/5 \)
\( tanθ=sinθ/cosθ=(4/5)/(3/5)=±4/3 \)
Answer:
±4/3
\( If tan θ = 2 and cos θ > 0, find sin θ and cos θ. \)