P2.6 Trigonometric identities and equations

Where do the two identities come from?

Draw a circle of radius 1 centred at the origin. A radius turned through an angle \(\theta\) anticlockwise from the positive \(x\)-axis meets the circle at a point \(P\). For any angle, \(\cos\theta\) is defined as the \(x\)-coordinate of \(P\) and \(\sin\theta\) as its \(y\)-coordinate. For acute angles this agrees with SOH CAH TOA, because the hypotenuse is 1; for larger angles it is what extends sin and cos to every angle.

• Pythagoras. \(P\) is 1 unit from \(O\), so \(x^2 + y^2 = 1\), which is \(\sin^2\theta + \cos^2\theta \equiv 1\). Here \(\sin^2\theta\) means \((\sin\theta)^2\).
• Gradient. The gradient of \(OP\) is \(\frac{y}{x}\), which is the definition of \(\tan\theta\), so \(\tan\theta \equiv \frac{\sin\theta}{\cos\theta}\). It is undefined where \(\cos\theta = 0\) (at \(90^\circ\), \(270^\circ\), …).
• The sign \(\equiv\) means "identically equal": true for every value of \(\theta\), not just for particular solutions.
• The Pythagorean identity rearranges into two forms that you will use constantly: \(\sin^2\theta \equiv 1 - \cos^2\theta\) and \(\cos^2\theta \equiv 1 - \sin^2\theta\).

How do I find exact values from one known ratio?

If you know one ratio, \(\sin^2\theta + \cos^2\theta = 1\) gives the size of the other, and the quadrant of \(\theta\) decides its sign. The signs follow from the unit circle: in the first quadrant (0° to 90°) all three are positive; in the second (90° to 180°) only sin is positive; in the third (180° to 270°) only tan; in the fourth (270° to 360°) only cos. Then \(\tan\theta = \sin\theta \div \cos\theta\). "Exact" means surds and fractions, so rationalise the denominator if asked for simplest form.

Given \(\tan\theta = k\) instead, write \(\sin\theta = k\cos\theta\) and substitute into the Pythagorean identity. A right-angled triangle with opposite side \(k\) and adjacent side 1 is an equally good method, as long as you attach the signs afterwards.

How do I prove a trigonometric identity?

An identity is a statement that must be true for every angle, so checking one or two values proves nothing: they could agree by chance. A proof is a chain of equal expressions that starts at one side and ends at the other.

• Start with the more complicated side and transform it into the other side. Write "LHS =" at the start and finish with "= RHS".
• Useful first moves: replace \(\tan\theta\) with \(\frac{\sin\theta}{\cos\theta}\); combine fractions over a common denominator; expand brackets; look for \(\sin^2\theta + \cos^2\theta\) and replace it with 1.
• Factorising helps too: \(a^2 - b^2 = (a - b)(a + b)\) is common with fourth powers.
• Do not start from the statement you are trying to prove and manipulate both sides: that assumes the result.

How do I find every solution of sin θ = k, cos θ = k or tan θ = k?

The calculator's inverse function returns one angle, the principal value \(\alpha\): \(\sin^{-1}\) and \(\tan^{-1}\) give an angle from \(-90^\circ\) to \(90^\circ\), and \(\cos^{-1}\) gives one from \(0^\circ\) to \(180^\circ\). That value may not even be in your interval. The graphs repeat and are symmetric, which gives a second angle in each cycle:

• \(\sin\theta = k\): \(\alpha\) and \(180^\circ - \alpha\) (the sine curve is symmetric about \(90^\circ\)).
• \(\cos\theta = k\): \(\alpha\) and \(360^\circ - \alpha\), equivalently \(-\alpha\) (the cosine curve is symmetric about \(0^\circ\)).
• \(\tan\theta = k\): \(\alpha\) and \(\alpha + 180^\circ\) (the tangent repeats every \(180^\circ\)).

Then add or subtract the period (\(360^\circ\) for sin and cos, \(180^\circ\) for tan) until every solution in the interval has been listed. If \(|k| > 1\), \(\sin\theta = k\) and \(\cos\theta = k\) have no solutions; \(\tan\theta = k\) always has solutions. Give degrees to 1 decimal place unless told otherwise, but carry the full calculator value through any later working.

These exact values must be known. In radians the angles are \(0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\) and \(\frac{\pi}{2}\).

What do I do when an equation contains both sin and cos?

There are three common shapes, and each has its own safe move.

• Linear in both, such as \(a\sin\theta = b\cos\theta\): divide both sides by \(\cos\theta\) to get \(\tan\theta = \frac{b}{a}\). This is safe, because \(\cos\theta = 0\) would force \(\sin\theta = 0\) as well, which never happens at the same angle.
• A common factor, such as \(\sin\theta\cos\theta = k\sin\theta\): do not divide by \(\sin\theta\). Bring everything to one side and factorise: \(\sin\theta(\cos\theta - k) = 0\). Then solve each factor \(= 0\).
• Squares of one, first powers of the other: use \(\sin^2\theta + \cos^2\theta = 1\) to reach a quadratic in one function (see the last teaching section).

How do I solve sin(nθ + α) = k in a given interval?

Let \(u = n\theta + \alpha\). The graph of \(\sin(n\theta + \alpha)\) has period \(\frac{360^\circ}{n}\), so there are about \(n\) times as many solutions in the interval as for \(\sin\theta = k\). The method has three steps:

1. Transform the interval. Apply the same operations to the end points: \(0^\circ \le \theta < 360^\circ\) becomes \(\alpha \le u < 360n^\circ + \alpha\).
2. Solve for \(u\) in the new interval exactly as before, adding \(360^\circ\) until you pass the upper limit (and checking below the lower limit too).
3. Convert back: \(\theta = \frac{u - \alpha}{n}\) for every value of \(u\).

Changing the interval first is what stops you missing solutions; dividing by \(n\) at the end, using unrounded values, keeps the answers accurate.

How does the same method work in radians?

If the interval is written with \(\pi\) (such as \(0 \le x < 2\pi\)), the answers must be in radians, so set the calculator to radian mode. The rules are the same with \(180^\circ\) replaced by \(\pi\): sin gives \(\alpha\) and \(\pi - \alpha\); cos gives \(\alpha\) and \(2\pi - \alpha\); tan gives \(\alpha\) and \(\alpha + \pi\); the period of sin and cos is \(2\pi\).

Exact answers are multiples of \(\pi\), from the exact-value table; other answers are usually given to 3 significant figures. For \(\sin(nx + \alpha) = k\), transform the interval exactly as before. It helps to write the end points with a common denominator, e.g. \(\frac{\pi}{4} \le 3x + \frac{\pi}{4} \le \frac{13\pi}{4}\).

How do I solve a quadratic equation in sin, cos or tan?

An equation such as \(2\sin^2\theta - \sin\theta - 1 = 0\) is a quadratic in \(\sin\theta\). Factorise (or use the formula) as if \(\sin\theta\) were a single letter \(s\). If the equation mixes \(\sin^2\theta\) with \(\cos\theta\), or \(\cos^2\theta\) with \(\sin\theta\), first replace the square using \(\sin^2\theta = 1 - \cos^2\theta\) or \(\cos^2\theta = 1 - \sin^2\theta\), so that only one function remains.

Each root gives a basic equation of the kind solved above. Reject any value of sin or cos outside \(-1\) to \(1\), and say why. Expect up to four solutions in a \(360^\circ\) interval: two from each root.

Exam tips for P2.6

Common misconceptions

How this topic connects

Trigonometry in P1

Radians, the exact values and the graphs of sin, cos and tan with their transformations are the starting point here. The ambiguous case of the sine rule is the equation \(\sin\theta = k\) with its two solutions \(\alpha\) and \(180^\circ - \alpha\).

Trigonometry in P3

sec, cosec and cot, the identities \(1 + \tan^2\theta \equiv \sec^2\theta\) and \(1 + \cot^2\theta \equiv \csc^2\theta\), compound and double angle formulae and \(R\cos(\theta - \alpha)\) all reduce equations to the forms solved on this page. Differentiating and integrating sin and cos (P3, P4) uses radians, and \(\sin^2 t + \cos^2 t = 1\) eliminates the parameter from curves such as \(x = 3\cos t,\ y = 2\sin t\).

Mechanics and IAL Physics

Resolving forces leads to equations such as \(\tan\theta = 0.75\) or \(\sin\theta = k\); oscillations and waves use \(\sin(\omega t + \varphi) = k\), solved for \(t\) by the interval method on this page.

Key formulae

Check your understanding

Specification checklist

• (P2 6.1) I know that \(\tan\theta = \sin\theta / \cos\theta\) and \(\sin^2\theta + \cos^2\theta = 1\), and can explain both from the unit circle.
• (P2 6.1) Given one of sin, cos or tan and the quadrant of the angle, I can find the exact values of the other two.
• (P2 6.1) I can prove simple identities by transforming one side into the other, using the two identities, common denominators and factorising.
• (P2 6.2) I can find all the solutions of \(\sin\theta = k\), \(\cos\theta = k\) and \(\tan\theta = k\) in a given interval, in degrees or in radians, exactly where possible.
• (P2 6.2) I can solve equations containing both sin and cos, by dividing to obtain tan or by factorising, without losing solutions.
• (P2 6.2) I can solve equations of the form \(\sin(n\theta + \alpha) = k\) (and the cos and tan versions) by first transforming the interval.
• (P2 6.2) I can solve quadratic equations in \(\sin\theta\), \(\cos\theta\) or \(\tan\theta\), using \(\sin^2\theta + \cos^2\theta = 1\) where needed, and reject impossible roots with a reason.