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MHT-CET Maths · Formula sheet

Trigonometry - II formulas

6 formulas and 14 common traps for MHT-CET Maths Trigonometry - II, grouped by subtopic.

Full notes with worked examples

Compound Angles and Conditional Identities

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The Compound-Angle Formulas and a sin x + b cos x

Compound angles

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B \qquad \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B \qquad \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}

Fixed Angle Sums and the Tangent Formula

Clearing the fraction

tan⁡(A+B)=k  ⟺  tan⁡A+tan⁡B=k (1−tan⁡Atan⁡B)\tan(A+B)=k \iff \tan A+\tan B = k\,(1-\tan A\tan B)

Common traps

Keeping the plus sign in cos(A + B)

cos(A + B) = cos A cos B − sin A sin B. The sine formula keeps the sign of the angle sum; the cosine formula reverses it.

Taking a + b as the maximum of a sin x + b cos x

The two terms never peak together. The maximum is a2+b2\sqrt{a^2 + b^2}, not a+ba + b and not a2+b2a^2 + b^2.

Forgetting that cot is the reciprocal

cot B − cot A = (tan A − tan B)/(tan A tan B). Given tan A − tan B = x and cot B − cot A = y, the product tan A tan B is x/y, and cot(A − B) = 1/x + 1/y.

Reading 225° as a new case

tan 225° = tan 45° = 1, so A + B = 225° gives exactly the same identity as A + B = 45°: the product of (1 + tan A)(1 + tan B) is 2.

Dropping a sign in the triple rearrangement

tan 3A(1 − tan 2A tan A) = tan 2A + tan A, so tan 3A − tan 2A − tan A = +tan A tan 2A tan 3A. The negative of it is printed as an option.

Double, Triple and Half Angles

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Double and Half Angles, and the Sign of a Half Angle

Double and half angles

cos⁡2A=2cos⁡2A−1=1−2sin⁡2Acos⁡x2=±1+cos⁡x2tan⁡x2=1−cos⁡xsin⁡x\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \qquad \cos\frac{x}{2}=\pm\sqrt{\frac{1+\cos x}{2}} \qquad \tan\frac{x}{2}=\frac{1-\cos x}{\sin x}

Triple Angles, Standard Values and Pairing

Triple angles

sin⁡3A=3sin⁡A−4sin⁡3Acos⁡3A=4cos⁡3A−3cos⁡Atan⁡3A=3tan⁡A−tan⁡3A1−3tan⁡2A\sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A \qquad \tan 3A=\frac{3\tan A-\tan^3 A}{1-3\tan^2 A}

Common traps

Taking the sign from the quadrant of x

With x in the third quadrant, cos x is negative, but x/2 is in the second quadrant and so is cos(x/2) — negative too. With x in the fourth, x/2 is in the second again. Always halve the interval first.

Using the wrong form of cos 2A

2cos²A − 1 and 1 − 2sin²A are both right; the choice decides whether the question collapses. For cos 2θ from sin²θ, use 1 − 2sin²θ.

The negative root of tan(π/8)

tan(π/8) solves t² + 2t − 1 = 0, whose roots are √2 − 1 and −1 − √2. π/8 is acute, so only the positive root is right; the other is an option.

Multiplying out four brackets

(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8) pairs into (1 − cos²π/8)(1 − cos²3π/8) = sin²(π/8) cos²(π/8) = 1/8. Expanding all four brackets is slow and error-prone.

Mixing up sin 18° and cos 36°

sin 18° = (√5 − 1)/4 and cos 36° = (√5 + 1)/4. They differ only in one sign, and both are printed as options.

Sum-to-Product and Product Formulas

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Sums to Products, and Ratio Conditions

Sum to product

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2} \qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}

Products to Sums, and the Two Square-Difference Identities

Product to sum

2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)2\cos A\cos B=\cos(A-B)+\cos(A+B) \qquad \cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B)

Common traps

The minus sign in cos C − cos D

cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2). Without the minus, cos 20° − cos 110° comes out negative when it is positive.

Inverting the ratio in componendo and dividendo

3 sin α = 5 sin β means sin α/sin β = 5/3, so the ratio is (5 + 3)/(5 − 3) = 4, not (3 + 5)/(3 − 5) = −4. Check which sine is larger first.

Using the sine pair for a cosine-minus-sine

cos²A − sin²B = cos(A + B) cos(A − B), but sin²A − sin²B = sin(A + B) sin(A − B). Mixing them gives cos 60° sin 36° instead of cos 60° cos 36°, and that value is printed too.

Using 18° where the value needs 36°

The question gives sin 18° = (√5 − 1)/4, but the answer needs cos 36° = 1 − 2sin²18° = (√5 + 1)/4. Quoting the given value unchanged gives (√5 − 1)/8, an option.

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