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CDS Mathematics · Trigonometric Ratios and Identities

Power Identities & Given Sums

When a question gives sin θ + cos θ, or a sin θ + b cos θ, square it: the square hands you sin θ cos θ, and sin θ cos θ is what every higher power and every partner expression is built from.

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Why this matters

Seventeen PYQs, every one MODERATE — the most uniform page in the chapter. Nothing here needs the angle itself; the whole page runs on one product, sin θ cos θ, and two power identities. Learn those, and these are fast, reliable marks.

Concept 1 of 3: Square the given sum to get sin θ cos θ

(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta, because the squares add to 11. So a given value of the sum is really a given value of the product — and sin⁡θ−cos⁡θ\sin\theta - \cos\theta, tan⁡θ+cot⁡θ\tan\theta + \cot\theta and sec⁡θ+cosec⁡θ\sec\theta + \operatorname{cosec}\theta are all simple functions of that product.

Definition

Write p=sin⁡θcos⁡θp = \sin\theta\cos\theta. If sin⁡θ+cos⁡θ=k\sin\theta + \cos\theta = k, then p=k2−12p = \dfrac{k^2 - 1}{2}, and:

  • (sin⁡θ−cos⁡θ)2=1−2p=2−k2(\sin\theta - \cos\theta)^2 = 1 - 2p = 2 - k^2;
  • tan⁡θ+cot⁡θ=1p\tan\theta + \cot\theta = \dfrac{1}{p};
  • sec⁡θ+cosec⁡θ=sin⁡θ+cos⁡θp=kp\sec\theta + \operatorname{cosec}\theta = \dfrac{\sin\theta + \cos\theta}{p} = \dfrac{k}{p};
  • cos⁡2θ−sin⁡2θ=(cos⁡θ+sin⁡θ)(cos⁡θ−sin⁡θ)\cos^2\theta - \sin^2\theta = (\cos\theta + \sin\theta)(\cos\theta - \sin\theta).

The sign of sin⁡θ−cos⁡θ\sin\theta - \cos\theta depends on whether θ\theta is above or below 45∘45^\circ.

The product from the sum

sin⁡θ+cos⁡θ=k  ⇒  sin⁡θcos⁡θ=k2−12,tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\sin\theta + \cos\theta = k \;\Rightarrow\; \sin\theta\cos\theta = \frac{k^2 - 1}{2}, \quad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}

Worked example

If sin⁡θ+cos⁡θ=75\sin\theta + \cos\theta = \dfrac{7}{5}, find tan⁡θ+cot⁡θ\tan\theta + \cot\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2016 · CDS (II) 2016 — Elementary Mathematics · Q71Moderate

Example 1 · Trigonometric Ratios and Identities · Power Identities and Given Sums

If sin⁡θ+cos⁡θ=72\sin\theta + \cos\theta = \frac{\sqrt{7}}{2}, then what is sin⁡θ−cos⁡θ\sin\theta - \cos\theta equal to ?

Squaring loses the sign

From (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 you get ±\pm. Above 45∘45^\circ the sine is larger and the difference is positive; below, it is negative. When the stem gives no range, both signs occur, and the paper will print only one of them.
Drill 7 more on square the given sum to get sin θ cos θ

Concept 2 of 3: sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ

Cube or square sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 and collect terms. What falls out is that every symmetric power sum is 11 minus a multiple of sin⁡2θcos⁡2θ\sin^2\theta\cos^2\theta. Two of these are worth memorising outright.

Definition

With p=sin⁡θcos⁡θp = \sin\theta\cos\theta:

  • sin⁡4θ+cos⁡4θ=1−2p2\sin^4\theta + \cos^4\theta = 1 - 2p^2
  • sin⁡6θ+cos⁡6θ=1−3p2\sin^6\theta + \cos^6\theta = 1 - 3p^2
  • sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(1−p)\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - p)

A long expression in these powers often reduces to a constant: the p2p^2 terms cancel, and the answer does not depend on θ\theta at all.

Power identities

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ,sin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta, \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta

Worked example

Find 3(sin⁡4θ+cos⁡4θ)−2(sin⁡6θ+cos⁡6θ)3(\sin^4\theta + \cos^4\theta) - 2(\sin^6\theta + \cos^6\theta).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q58Moderate

Example 2 · Trigonometric Ratios and Identities · Power Identities and Given Sums

What is 2sin⁡6θ+2cos⁡6θ−3sin⁡4θ−3cos⁡4θ2\sin^6\theta + 2\cos^6\theta - 3\sin^4\theta - 3\cos^4\theta equal to ?

It is minus, not plus

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta. A statement with +2+2 is planted in identity questions and looks right at a glance. Check it at 45∘45^\circ: the true value is 12\dfrac12, and the plus version gives 32\dfrac32.
Drill 5 more on sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ

Concept 3 of 3: a sin θ + b cos θ and its partner a cos θ − b sin θ

Square asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta and acos⁡θ−bsin⁡θa\cos\theta - b\sin\theta and add: the cross terms ±2absin⁡θcos⁡θ\pm 2ab\sin\theta\cos\theta cancel, and what is left is a2+b2a^2 + b^2. So knowing one of the pair fixes the square of the other — no need to find θ\theta.

Definition

  • (asin⁡θ+bcos⁡θ)2+(acos⁡θ−bsin⁡θ)2=a2+b2(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2.
  • So if asin⁡θ+bcos⁡θ=ca\sin\theta + b\cos\theta = c, then (acos⁡θ−bsin⁡θ)2=a2+b2−c2(a\cos\theta - b\sin\theta)^2 = a^2 + b^2 - c^2.
  • The partner is found up to sign; both signs usually occur for different angles. When the sum reaches its maximum a2+b2\sqrt{a^2 + b^2}, the partner is 00 — see the maximum page.

Partner identity

(asin⁡θ+bcos⁡θ)2+(acos⁡θ−bsin⁡θ)2=a2+b2(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2

Worked example

If 4sin⁡θ+3cos⁡θ=24\sin\theta + 3\cos\theta = 2, find (4cos⁡θ−3sin⁡θ)2(4\cos\theta - 3\sin\theta)^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q69Moderate

Example 3 · Trigonometric Ratios and Identities · Power Identities and Given Sums

If 3sin⁡θ+5cos⁡θ=43 \sin \theta + 5 \cos \theta = 4, then what is the value of (3cos⁡θ−5sin⁡θ)2(3 \cos \theta - 5 \sin \theta)^2 ?

When only one sign is printed, it is not the only answer

3sin⁡θ+5cos⁡θ=53\sin\theta + 5\cos\theta = 5 gives 5sin⁡θ−3cos⁡θ=±35\sin\theta - 3\cos\theta = \pm 3, and both signs genuinely happen (θ=0∘\theta = 0^\circ gives −3-3). The paper printed only −3-3. Choose the value offered; do not conclude your +3+3 was wrong.
Drill 2 more on a sin θ + b cos θ and its partner a cos θ − b sin θ

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Square the given sum to get sin θ cos θ

    The product from the sum

    sin⁡θ+cos⁡θ=k  ⇒  sin⁡θcos⁡θ=k2−12,tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\sin\theta + \cos\theta = k \;\Rightarrow\; \sin\theta\cos\theta = \frac{k^2 - 1}{2}, \quad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}
  • sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ

    Power identities

    sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ,sin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta, \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta
  • a sin θ + b cos θ and its partner a cos θ − b sin θ

    Partner identity

    (asin⁡θ+bcos⁡θ)2+(acos⁡θ−bsin⁡θ)2=a2+b2(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2

Watch out for (3)

Test yourself on Trigonometric Ratios and Identities

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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