Two identities do almost all the work at AS: sin²θ + cos²θ ≡ 1 and tan θ ≡ sin θ / cos θ. You use them to simplify expressions, find the other ratios from one given ratio, prove identities, and convert an equation that mixes sin, cos and tan into a quadratic in a single trig function.
The two key identities
- sin²θ + cos²θ ≡ 1
- sin²θ ≡ 1 − cos²θ and cos²θ ≡ 1 − sin²θ
- tan θ ≡ sin θ / cos θ
Typical route: write tan as a fraction, clear the denominator, then use sin²θ + cos²θ = 1 to leave one function.
What are the two trig identities?
The first comes from Pythagoras on the unit circle: sin²θ + cos²θ ≡ 1. It rearranges both ways, so you can replace sin²θ with 1 − cos²θ or cos²θ with 1 − sin²θ. The second, tan θ ≡ sin θ / cos θ, lets you turn any tan into a fraction, which you can then clear by multiplying through.
Nearly every question uses both together.
How do you find cos θ and tan θ from sin θ?
Use the identity to get the second ratio, then the fraction to get the third: cos θ = ±√(1 − sin²θ), and tan θ = sin θ / cos θ.
The ± is the whole difficulty. Decide the sign from the interval in the question by picturing the graph. The interval is given precisely to tell you which sign to keep, so never drop the ± without checking.
How do you prove a trig identity?
Start on one side, usually the more complicated one, and work towards the other. You may not move terms across the ≡ sign. The usual toolkit is: write tan as a fraction, combine over a common denominator, factorise (often a difference of two squares), and replace 1 with sin²θ + cos²θ when a lone 1 needs breaking up.
That last move is the one students most often miss. It turns a constant into something that can be factorised or cancelled.
Show that an equation becomes a quadratic
The classic exam question gives an equation mixing sin, cos and tan, and asks you to show it can be written as a quadratic in one function. The route is the same each time: write tan as sin / cos, multiply through by cos, replace sin² with 1 − cos², and collect everything on one side.
Look at which function survives. If clearing the fraction leaves sin² and cos terms, converting the sin² gives a quadratic in cos. A root such as cos x = −2 must be rejected, and saying so explicitly earns a mark.
Dividing by cos²x
If every term in an equation is a product of two of sin x and cos x, such as sin²x, sin x cos x and cos²x, then dividing every term by cos²x turns it into a quadratic in tan x, because sin²x / cos²x = tan²x and sin x cos x / cos²x = tan x.
Collect everything on one side before dividing, so the right-hand side is zero.
Worked examples
Example 1. Simplify tan θ + 1 / tan θ, giving your answer as a single fraction.
- Write both as fractions: sin θ / cos θ + cos θ / sin θ.
- Common denominator: (sin²θ + cos²θ) / (sin θ cos θ).
- The numerator is exactly 1.
Answer: 1 / (sin θ cos θ).
Marks: M1 for the common denominator, A1 for applying the identity.
Example 2. Given that sin θ = 3/5 and θ is obtuse, find the exact values of cos θ and tan θ.
- cos²θ = 1 − sin²θ = 1 − 9/25 = 16/25.
- cos θ = ±4/5.
- θ is obtuse, so the cosine curve is below the axis: cos θ = −4/5.
- tan θ = sin θ / cos θ = (3/5) / (−4/5) = −3/4.
Answer: cos θ = −4/5, tan θ = −3/4.
Marks: M1 for the identity, A1 for the sign, A1 for tan θ.
Example 3. Prove that tan²θ − sin²θ ≡ tan²θ sin²θ.
- Start with the left-hand side and write tan as a fraction: sin²θ / cos²θ − sin²θ.
- Take out the common factor sin²θ: sin²θ(1 / cos²θ − 1).
- Combine the bracket: sin²θ × (1 − cos²θ) / cos²θ.
- Use 1 − cos²θ = sin²θ: sin²θ × sin²θ / cos²θ.
- Since sin²θ / cos²θ = tan²θ, this is tan²θ sin²θ, the right-hand side.
Marks: M1 for tan as a fraction, M1 for factorising, A1 for the identity and conclusion.
Example 4. Show that the equation 2 tan x sin x = 3 can be written as 2 cos²x + 3 cos x − 2 = 0.
- tan as a fraction: 2 × (sin x / cos x) × sin x = 3, so 2 sin²x / cos x = 3.
- Multiply through by cos x: 2 sin²x = 3 cos x.
- Use sin²x = 1 − cos²x: 2 − 2 cos²x = 3 cos x.
- Collect on one side: 2 cos²x + 3 cos x − 2 = 0, as required.
Marks: M1 for tan as a fraction, M1 for clearing the fraction, M1 for the identity, A1 for the quadratic.
Common mistakes
- Dropping the ± after a square root. cos θ = ±√(1 − sin²θ). Use the interval to choose the sign.
- Moving terms across the ≡ sign in a proof. Work from one side to the other without rearranging the identity itself.
- Writing sin²θ as sin θ². sin²θ means (sin θ)², the sine squared, not the sine of θ².
- Dividing by cos²x before collecting terms. Get everything on one side first, so each term divides cleanly.
- Not rejecting an impossible root. sin x and cos x lie between −1 and 1. State that a root like cos x = −2 is rejected, and why.
How the trig identities are examined
In Edexcel International A-Level, these identities are Pure 2 content (paper WMA12), in the trigonometric identities and equations chapter. Edexcel A-Level (9MA0) students meet them in the first year of Pure. Questions typically ask you to:
- find the exact values of the other ratios from one given ratio and an interval
- prove a short identity
- show that an equation can be written as a quadratic in sin x, cos x or tan x
- factorise that quadratic and say which root must be rejected
Exam-style questions in these formats, with mark schemes, are in the Pure 2 trigonometric identities and equations exam questions. Use the trig identities and equations worksheet or the trigonometry revision engine for more.
The next step is solving trig equations over an interval. In Pure 3 the same ideas extend to sec, cosec and cot.
Frequently asked questions
Why is sin²θ + cos²θ = 1?
On a unit circle, a point at angle θ has coordinates (cos θ, sin θ). Pythagoras on the right-angled triangle with hypotenuse 1 gives cos²θ + sin²θ = 1.
When should I divide by cos²x?
When every term is degree 2 in sin x and cos x, such as sin²x, sin x cos x and cos²x. Dividing by cos²x then gives a quadratic in tan x.
What is the difference between an identity and an equation?
An identity, written with ≡, is true for every value of θ. An equation is only true for particular values, which you have to find.