The value of (sin²θ − 2sin⁴θ)⁄(2cos⁴θ + cos²θ) at θ = 45° is:
- (a)3
- (b)0
- (c)2
- (d)1
Answer
Why
Correct — B. Evaluate numerator and denominator separately at θ = 45°; the numerator collapses.
At 45°, sin θ = cos θ = 1⁄√2, so sin²θ = cos²θ = 1⁄2 and sin⁴θ = cos⁴θ = 1⁄4.
Numerator = sin²θ − 2sin⁴θ
= 1⁄2 − 2 × 1⁄4 = 1⁄2 − 1⁄2 = 0
Denominator = 2cos⁴θ + cos²θ
= 2 × 1⁄4 + 1⁄2 = 1⁄2 + 1⁄2 = 1
Value = 0 ÷ 1 = 0 → option (b).
Why it vanishes: the numerator factors as sin²θ(1 − 2sin²θ) = sin²θ · cos 2θ, and cos 90° = 0.
Why the others are wrong
- (a)3 — The denominator works out to exactly 1, so the value of the fraction equals its numerator, and the numerator is 1⁄2 − 1⁄2 = 0. Nothing in the expression yields 3.
- (c)2 — 2 is a coefficient inside the expression, not its value. With the denominator equal to 1, the answer is whatever the numerator is, and at 45° that is zero.
- (d)1 — 1 is the value of the denominator, since 2cos⁴45° + cos²45° = 1⁄2 + 1⁄2. The fraction is 0 ÷ 1, so 1 is the divisor and not the answer.
Concept
Substituting a special angle is the fastest route when the question fixes θ, but it pays to notice the structure first.
At 45° everything is symmetric: sin 45° = cos 45° = 1⁄√2, so sin²θ and cos²θ are both 1⁄2, and both fourth powers are 1⁄4.
The numerator factors as sin²θ(1 − 2sin²θ) = sin²θ · cos 2θ, and cos 2θ is zero when θ = 45°. That is why the whole fraction is zero, not a coincidence of arithmetic.
Once the numerator is zero, the denominator matters only in that it must not also be zero — here it is 1.
The stem for this item is a question image rather than text.
It asks for the value of (sin²θ − 2sin⁴θ) ÷ (2cos⁴θ + cos²θ) at θ = 45°.
Key facts
- sin 45° = cos 45° = 1⁄√2, so sin²45° = cos²45° = 1⁄2 and sin⁴45° = cos⁴45° = 1⁄4.
- 1 − 2sin²θ = cos 2θ, which is zero at θ = 45°.
- The denominator 2cos⁴45° + cos²45° equals 1, so the fraction reduces to its numerator.
Study next
Common traps
- Writing sin⁴45° as 1⁄2 instead of 1⁄4 by forgetting to square sin²45°.
- Reporting the denominator's value of 1 once the numerator turns out to be zero.
- Assuming a fraction cannot evaluate to zero and hunting for a slip that is not there.
Trigonometry here appears both as substitution and as identity work.
18 Sep 2024, 09:00, Quant Q.7 asks at what angle sin θ and cos θ take the same value, while 23 Sep 2024, 12:30, Quant Q.18 gives cos A + cos²A = 1 and asks for sin²A + sin⁴A.
Related PYQs
No directly related past PYQ was found.