Exact value of cos 20 deg + 2sin^2 (55 deg) - sqrt(2) * sin 65 deg * k 8. (A) 1 (B) 1/(sqrt(2)) (C) sqrt(2) (D) zero

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Published July 10, 2025
Mathematics
Trigonometry
Standard-angle identities
Sum & difference formulae

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Detailed Explanation

Key tools you must know

  1. Pythagorean identity (double-angle form)
2sin⁡2x=1−cos⁡(2x)2\sin^2x = 1-\cos(2x)
  1. Co-function identity
sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta) = \cos\theta
  1. Angle–sum formula for sine
sin⁡(A+B)=sin⁡A cos⁡B+cos⁡A sin⁡B\sin(A+B) = \sin A\,\cos B + \cos A\,\sin B
  1. Exact values at 45°
sin⁡45°=cos⁡45°=22\sin45° = \cos45° = \frac{\sqrt2}{2}

Student’s logical roadmap

  1. Convert the squared sine term: Replace 2sin⁡255°2\sin^2 55° by 1−cos⁡110°1-\cos110°. Because 110°=180°−70°110° = 180°-70°, we turn cos⁡110°\cos110° into −cos⁡70°-\cos70°. So the whole piece simplifies to 1+cos⁡70°1+\cos70°.

  2. Expand the 65° sine: Recognise 65°=45°+20°65° = 45° + 20°. Use the sum formula and exact sin⁡45°,cos⁡45°\sin45°, \cos45° values, then multiply by 2\sqrt2 so the pesky 2\sqrt2 cancels neatly.

  3. Collect like terms: You will get a +cos⁡20°+\cos20° from the first chunk and a −cos⁡20°-\cos20° from the expanded chunk – they cancel. What remains is 1+cos⁡70°−sin⁡20°1 + \cos70° - \sin20°.

  4. Use the co-function rule: Replace sin⁡20°\sin20° with cos⁡70°\cos70°. They also cancel, leaving 1.

  5. Match to the options. The only choice equal to 1 is option (A).

Simple Explanation (ELI5)

What’s the question?

We have to find the exact (not decimal-approximate) value of this funny looking mix of cos and sin angles:

cos 20°  +  2·sin²55°  –  √2·sin65°

How do we tame it?

  1. Turn big things into small, friendly ones. We know some special angles like 45°, 60°, 30°. Notice 55° and 65° are close to 45° plus or minus 10° or 20°.
  2. Use identities as Lego bricks. Identities are like Lego pieces that let you replace a messy block with a simpler, nicer-looking one that fits exactly.
    • 2sin⁡2x=1−cos⁡(2x)2\sin^2 x = 1-\cos(2x) (Pythagorean double-angle)
    • sin⁡(45°+θ)=22(cos⁡θ+sin⁡θ)\sin(45°+\theta) = \frac{\sqrt2}{2}\big(\cos\theta+\sin\theta\big) (sum formula)
    • sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta)=\cos\theta (co-function rule)
  3. Watch pieces cancel out. When you substitute, terms like +cos⁡20°+\cos20° and −cos⁡20°-\cos20° magically disappear, leaving a very small remainder.
  4. End up with a single number. All the angle stuff vanishes and we’re left with exactly 1. No calculator needed if you trust the identities!

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Step-by-Step Solution

Step-by-step computation

  1. Rewrite the 2sin⁡22\sin^2 term
2sin⁡255°=1−cos⁡(2×55°)=1−cos⁡110°2\sin^2 55° = 1 - \cos(2\times55°) = 1 - \cos110°

But cos⁡110°=cos⁡(180°−70°)=−cos⁡70°\cos110° = \cos(180° - 70°) = -\cos70°, so

2sin⁡255°=1+cos⁡70°2\sin^2 55° = 1 + \cos70°
  1. Rewrite the 2 sin⁡65°\sqrt2\,\sin65° term

Notice 65°=45°+20°65° = 45° + 20°. Using the sum formula:

sin⁡65°=sin⁡(45°+20°)=sin⁡45°cos⁡20°+cos⁡45°sin⁡20°\sin65° = \sin(45° + 20°) = \sin45°\cos20° + \cos45°\sin20°

Because sin⁡45°=cos⁡45°=22\sin45° = \cos45° = \dfrac{\sqrt2}{2},

sin⁡65°=22(cos⁡20°+sin⁡20°)\sin65° = \frac{\sqrt2}{2}(\cos20° + \sin20°)

Multiply by 2\sqrt2:

2 sin⁡65°=2×22(cos⁡20°+sin⁡20°)=cos⁡20°+sin⁡20°\sqrt2\,\sin65° = \sqrt2 \times \frac{\sqrt2}{2}(\cos20° + \sin20°) = \cos20° + \sin20°
  1. Assemble the original expression

Original expression:

cos⁡20°+2sin⁡255°−2 sin⁡65°\cos20° + 2\sin^2 55° - \sqrt2\,\sin65°

Replace each part:

cos⁡20°+(1+cos⁡70°)−(cos⁡20°+sin⁡20°)\cos20° + \big(1 + \cos70°\big) - \big(\cos20° + \sin20°\big)
  1. Cancel like terms
cos⁡20°+1+cos⁡70°−cos⁡20°−sin⁡20°=1+cos⁡70°−sin⁡20°\color{gray}{\cos20°} + 1 + \cos70° - \color{gray}{\cos20°} - \sin20° = 1 + \cos70° - \sin20°
  1. Use the co-function identity (sin⁡20°=cos⁡(90°−20°)=cos⁡70°\sin20° = \cos(90°-20°) = \cos70°)
1+cos⁡70°−sin⁡20°=1+cos⁡70°−cos⁡70°=11 + \cos70° - \sin20° = 1 + \cos70° - \cos70° = 1
  1. Choose the correct option

Option (A) 1\boxed{1}

Examples

Example 1

In alternating-current circuits, expressing power as 2·sin² terms allows engineers to isolate the DC component.

Example 2

GPS signal processing uses angle-sum identities to simplify phase differences between satellites.

Example 3

Optics: Young’s double-slit interference relies on converting sin(A±B) to predict bright & dark fringes.

Example 4

In 3-D graphics, rotations often combine angles; breaking them with sum/difference formulas speeds computation.

Visual Representation

References

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