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NCERT Exemplar · Q49

Q.The value of sin⁡50∘−sin⁡70∘+sin⁡10∘\sin 50^\circ - \sin 70^\circ + \sin 10^\circ is equal to
(A) 11
(B) 00
(C) 12\dfrac{1}{2}
(D) 22

Rajasthan RbseMCQ· 1mImportance★★★★★est
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Regroup the terms strategically and apply the sum-to-product formula sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} to simplify. The expression equals 0.

The key insight here is recognizing that we have three sine terms at special angles (10°10°, 50°50°, 70°70°) that are related. Rather than computing each sine value separately, we should look for a way to combine them using trigonometric identities.

Notice that 10°+70°=80°10° + 70° = 80° and 50°50° sits somewhat in between. This suggests pairing sin⁡10°\sin 10° with sin⁡50°\sin 50° first, then dealing with the subtraction.

Let me rearrange the expression to make the pattern clearer:

sin⁡50°−sin⁡70°+sin⁡10°=(sin⁡50°+sin⁡10°)−sin⁡70°\sin 50° - \sin 70° + \sin 10° = (\sin 50° + \sin 10°) - \sin 70°

Now I'll work through this systematically.

1. Apply the sum-to-product formula to the first pair

The sum-to-product identity states:

sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

For sin⁡50°+sin⁡10°\sin 50° + \sin 10°:

sin⁡50°+sin⁡10°=2sin⁡(50°+10°2)cos⁡(50°−10°2)\sin 50° + \sin 10° = 2\sin\left(\frac{50° + 10°}{2}\right)\cos\left(\frac{50° - 10°}{2}\right)

=2sin⁡(30°)cos⁡(20°)= 2\sin(30°)\cos(20°)

2. Substitute the known value of sin⁡30°\sin 30°

We know that sin⁡30°=12\sin 30° = \frac{1}{2}, so:

2sin⁡(30°)cos⁡(20°)=2⋅12⋅cos⁡(20°)=cos⁡(20°)2\sin(30°)\cos(20°) = 2 \cdot \frac{1}{2} \cdot \cos(20°) = \cos(20°)

3. Rewrite the original expression

Our expression now becomes:

sin⁡50°−sin⁡70°+sin⁡10°=cos⁡(20°)−sin⁡70°\sin 50° - \sin 70° + \sin 10° = \cos(20°) - \sin 70°

4. Use the complementary angle relationship …

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