KBAT: Trig Function Reasoning

Q: How do I know whether to use a double-angle identity or the auxiliary-angle method?

Look at what the expression mixes. An equation with two different multiples of the same angle, such as cos⁡2x\cos 2xcos2x and sin⁡x\sin xsinx, needs a double-angle identity to bring everything into one angle before it can be solved. An expression that adds a sine term and a cosine term of the same angle, such as 3sin⁡x+4cos⁡x3\sin x+4\cos x3sinx+4cosx, needs the auxiliary-angle method, Rsin⁡(x+α)R\sin(x+\alpha)Rsin(x+α), which turns the sum into a single wave.

Q: Why is \sin x=2 rejected instead of solved?

Because sin⁡x\sin xsinx can never exceed 111 or fall below −1-1−1 for any real angle xxx. When solving a quadratic in sin⁡x\sin xsinx produces a root like 222, it is not a missing angle, it is an algebraic root with no matching angle at all, and must be explicitly rejected before the final answer is given.

Q: How do I find R and \alpha in R\sin(x+\alpha)?

Expand Rsin⁡(x+α)R\sin(x+\alpha)Rsin(x+α) using the addition formula to get Rcos⁡αsin⁡x+Rsin⁡αcos⁡xR\cos\alpha\sin x+R\sin\alpha\cos xRcosαsinx+Rsinαcosx, then compare coefficients with the original expression. Squaring and adding the two coefficient equations gives RRR; dividing them gives tan⁡α\tan\alphatanα, and hence α\alphaα, inside whichever range the question states, such as 0∘<α<90∘0^{\circ}<\alpha<90^{\circ}0∘<α<90∘.

Q: Why can't 3\sin x+4\cos x reach 3+4=7?

Because sin⁡x\sin xsinx and cos⁡x\cos xcosx cannot both equal 111 at the same angle xxx, when one is at its peak, the other is not. Writing the expression as Rsin⁡(x+α)R\sin(x+\alpha)Rsin(x+α) shows its true greatest value is R=32+42=5R=\sqrt{3^{2}+4^{2}}=5R=32+42=5, reached at one specific angle, which is smaller than the simple sum of the two coefficients.

What makes this a KBAT question

A routine trigonometric question already tells you the identity to use, or gives you an equation in one ratio ready to solve. A trigonometric-functions KBAT question mixes a double angle with a single angle in the same equation, so before anything can be solved you must choose an identity that rewrites both sides in one angle, and later in the same problem you may have to reason about a claimed value rather than only calculate one.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: the identity $\cos 2A=1-2\sin^{2}A$ and writing $a\sin x+b\cos x$ as $R\sin(x+\alpha)$ are both familiar Form 5 techniques, but here you decide which identity removes the mismatch, solve without losing a root, and then judge whether a stated maximum is even reachable. In Add Math this rewards students who understand what an identity does, rewrite an expression without changing its value, rather than students who only substitute into a memorised line.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[10 marks]

(a) Solve the equation $\cos 2x+5\sin x=3$ for $0^{\circ}\le x\le 360^{\circ}$ . (b) Express $f(x)=3\sin x+4\cos x$ in the form $R\sin(x+\alpha)$ , where $R>0$ and $0^{\circ}<\alpha<90^{\circ}$ .

Hence state the maximum and minimum values of $f(x)$ for $0^{\circ}\le x\le 360^{\circ}$ , and the corresponding value of $x$ in each case. (c) A student claims that the equation $3\sin x+4\cos x=6$ has a solution for $0^{\circ}\le x\le 360^{\circ}$ , reasoning that since $3$ and $4$ are both positive, their sum can reach $6$ .

Using your answer to (b), explain whether the student is correct.

Show worked solution

Understand. Part (a) asks for every angle $x$ in one full turn that satisfies an equation mixing $\cos 2x$ and $\sin x$ .

Part (b) asks for the largest and smallest values a different expression, $f(x)=3\sin x+4\cos x$ , can take over the same range, and the angle where each occurs. Part (c) tests whether a claimed value of $f(x)$ is actually possible.

Plan. For (a), rewrite $\cos 2x$ using $\cos 2A=1-2\sin^{2}A$ so the whole equation is in $\sin x$ , then solve the resulting quadratic and reject any root outside the range of sine.

For (b), expand $R\sin(x+\alpha)$ with the addition formula and compare coefficients with $3\sin x+4\cos x$ to find $R$ and $\alpha$ ; the maximum and minimum then follow from the range of $\sin(x+\alpha)$ . For (c), compare the claimed value with the maximum found in (b).

Execute and check, part (a). Replace $\cos 2x$ with $1-2\sin^{2}x$ :

1-2\sin^{2}x+5\sin x=3

Collect everything on one side and simplify:

2\sin^{2}x-5\sin x+2=0

Factorise:

(2\sin x-1)(\sin x-2)=0

So $\sin x=\tfrac{1}{2}$ or $\sin x=2$ . Since $-1\le\sin x\le 1$ for every angle, $\sin x=2$ is impossible and is rejected.

For $\sin x=\tfrac{1}{2}$ , the basic angle is $30^{\circ}$ ; sine is positive in the first and second quadrants, so

x=30^{\circ}\quad\text{or}\quad x=180^{\circ}-30^{\circ}=150^{\circ}

Execute and check, part (b). Expand the right side using $\sin(A+B)=\sin A\cos B+\cos A\sin B$ :

R\sin(x+\alpha)=R\cos\alpha\sin x+R\sin\alpha\cos x

Comparing this with $3\sin x+4\cos x$ gives $R\cos\alpha=3$ and $R\sin\alpha=4$ . Squaring and adding removes $\alpha$ :

R^{2}=3^{2}+4^{2}=25\ \Rightarrow\ R=5

Dividing the two equations gives $\tan\alpha=\tfrac{4}{3}$ , so $\alpha=53.13^{\circ}$ (to 2 d.p.), which fits $0^{\circ}<\alpha<90^{\circ}$ . So

f(x)=3\sin x+4\cos x=5\sin(x+53.13^{\circ})

As $x$ runs from $0^{\circ}$ to $360^{\circ}$ , $x+53.13^{\circ}$ runs from $53.13^{\circ}$ to $413.13^{\circ}$ , one full turn. The greatest value of $\sin(x+53.13^{\circ})$ is $1$ , reached once in that span at $x+53.13^{\circ}=90^{\circ}$ , giving $x=36.87^{\circ}$ ; the least value is $-1$ , reached at $x+53.13^{\circ}=270^{\circ}$ , giving $x=216.87^{\circ}$ .

So the maximum value of $f(x)$ is $5$ at $x=36.87^{\circ}$ , and the minimum value is $-5$ at $x=216.87^{\circ}$ .

Check. At $x=30^{\circ}$ : $\cos 60^{\circ}+5\sin 30^{\circ}=0.5+5(0.5)=3$ , which matches.

At $x=36.87^{\circ}$ : $\sin x=0.6$ and $\cos x=0.8$ , so $f(x)=3(0.6)+4(0.8)=1.8+3.2=5$ , confirming the maximum.

Part (c), reasoning. The student is not correct.

Part (b) shows that $f(x)=3\sin x+4\cos x$ can never exceed $R=5$ , for any angle $x$ , so $f(x)=6$ has no solution at all, not just none in $0^{\circ}\le x\le 360^{\circ}$ . The student's reasoning fails because it treats $3$ and $4$ as if they could both reach their full size at the same $x$ ; but $\sin x=1$ only at $x=90^{\circ}$ , where $\cos x=0$ , not $1$ .

The two terms cannot peak together, and the identity $\sin^{2}x+\cos^{2}x=1$ is exactly what stops them: the combined amplitude is bounded by $R=\sqrt{3^{2}+4^{2}}=5$ , not by the plain sum $3+4=7$ .

Finding a sensible first step

When a trigonometric equation or expression looks unfamiliar, name the mismatch before doing anything else. Two different multiples of the angle in the same equation, such as $2x$ and $x$ , call for a double-angle identity that puts everything in one angle.

Two different trig ratios added together, such as $3\sin x+4\cos x$ , call for the auxiliary-angle method, $R\sin(x+\alpha)$ , so the expression becomes a single wave with one maximum and one minimum. Once an equation is down to one ratio, solve it as an ordinary quadratic or linear equation, remembering that $\sin x$ and $\cos x$ can only ever lie between $-1$ and $1$ , any other 'solution' from the algebra must be rejected.

This last check is what turns an equation with an extra root into a fully correct answer.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a trigonometric-functions KBAT question a marker looks for:

The correct identity chosen to remove the mismatch, a double-angle identity for two multiples of the angle, the auxiliary-angle method for a sum of sine and cosine.
The substitution carried through correctly before any solving begins.
Any root outside the range of sine or cosine, such as $\sin x=2$ , explicitly rejected, not silently dropped.
The basic angle found from the positive ratio, then the correct quadrants used to list every solution in the stated range.
$R$ and $\alpha$ found by comparing coefficients, with $\alpha$ checked against the stated range such as $0^{\circ}<\alpha<90^{\circ}$ .
A reasoned answer in a 'show that' or 'explain whether' part, not just a numerical value.

How a teacher helps

Trigonometric identities are easy to state and easy to misuse, so our teachers spend the first few minutes of a lesson on naming the mismatch before reaching for a formula, a double angle beside a single angle, or two different ratios added together. We rehearse the auxiliary-angle method until writing $a\sin x+b\cos x$ as $R\sin(x+\alpha)$ is automatic, and we treat the range check on $\sin x$ and $\cos x$ as a habit, not an afterthought, so a rejected root is never missed.

We also practise the reasoning parts out loud, because a marker rewards a clear explanation of why a claim is true or false, not only a correct number. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

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Frequently asked questions

How do I know whether to use a double-angle identity or the auxiliary-angle method?

Look at what the expression mixes. An equation with two different multiples of the same angle, such as $\cos 2x$ and $\sin x$ , needs a double-angle identity to bring everything into one angle before it can be solved.

An expression that adds a sine term and a cosine term of the same angle, such as $3\sin x+4\cos x$ , needs the auxiliary-angle method, $R\sin(x+\alpha)$ , which turns the sum into a single wave.

Why is

\sin x=2

rejected instead of solved?

Because $\sin x$ can never exceed $1$ or fall below $-1$ for any real angle $x$ . When solving a quadratic in $\sin x$ produces a root like $2$ , it is not a missing angle, it is an algebraic root with no matching angle at all, and must be explicitly rejected before the final answer is given.

How do I find

R

and

\alpha

R\sin(x+\alpha)

Expand $R\sin(x+\alpha)$ using the addition formula to get $R\cos\alpha\sin x+R\sin\alpha\cos x$ , then compare coefficients with the original expression. Squaring and adding the two coefficient equations gives $R$ ; dividing them gives $\tan\alpha$ , and hence $\alpha$ , inside whichever range the question states, such as $0^{\circ}<\alpha<90^{\circ}$ .

Why can't

3\sin x+4\cos x

reach

3+4=7

Because $\sin x$ and $\cos x$ cannot both equal $1$ at the same angle $x$ , when one is at its peak, the other is not. Writing the expression as $R\sin(x+\alpha)$ shows its true greatest value is $R=\sqrt{3^{2}+4^{2}}=5$ , reached at one specific angle, which is smaller than the simple sum of the two coefficients.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Reasoning With Trigonometric Functions