Let p=sin⁡24∘1. Then what would cos⁡(24∘) be in terms of p?2. What would sin⁡(168∘)⋅sin⁡(−78∘) be in terms of p?

Part 1:Use the identitysin2A+cos2A=1cos2A=1−sin2Asocos224∘=1−sin2(24∘)cos224∘=1−p2cos⁡24∘=1−p2Here we only take positive value of square root because 24∘ is in I quaderent and cos ratio in quaderent I is positive.

Part 2sin⁡168∘⋅sin⁡(−78∘)=−2sin⁡168∘⋅sin⁡78∘2Using the formula 2sin⁡A⋅sin⁡B=cos⁡(A−B)−cos⁡(A+B)sin⁡(168∘)⋅sin⁡(−78∘)=−cos⁡90∘−cos⁡246∘2=cos⁡246∘2=cos⁡(270∘−24∘)2=−p2

Part 1:Use the identitysin2A+cos2A=1cos2A=1−sin2Asocos224∘=1−sin2(24∘)cos224∘=1−p2cos⁡24∘=1−p2Here we only take positive value of square root because 24∘ is in I quaderent and cos ratio in quaderent I is positive.

Part 2sin⁡168∘⋅sin⁡(−78∘)=−2sin⁡168∘⋅sin⁡78∘2Using the formula 2sin⁡A⋅sin⁡B=cos⁡(A−B)−cos⁡(A+B)sin⁡(168∘)⋅sin⁡(−78∘)=−cos⁡90∘−cos⁡246∘2=cos⁡246∘2=cos⁡(270∘−24∘)2=−p2

Find out the answer to "Let p=sin⁡24∘1. Then what would cos⁡(24∘) be in..."

Tagiuraoob

Answered question

2022-03-01

Let

p = {\sin 24}^{\circ}

1. Then what would

\cos (24^{\circ})

be in terms of p?
2. What would

\sin (168^{\circ}) \cdot \sin (- 78^{\circ})

be in terms of p?

Answer & Explanation

Warenberg56i

Beginner2022-03-02Added 8 answers

Part 1:
Use the identity

\sin^{2} A + \cos^{2} A = 1

\cos^{2} A = 1 - \sin^{2} A

\cos^{2} 24^{\circ} = 1 - \sin^{2} (24^{\circ})

\cos^{2} 24^{\circ} = 1 - p^{2}

{\cos 24}^{\circ} = \sqrt{1 - p^{2}}

Here we only take positive value of square root because

24^{\circ}

is in I quaderent and cos ratio in quaderent I is positive.

Ulgelmorgs

Beginner2022-03-03Added 8 answers

Part 2

{\sin 168}^{\circ} \cdot \sin (- 78^{\circ}) = - \frac{2 {\sin 168}^{\circ} \cdot {\sin 78}^{\circ}}{2}

Using the formula

2 \sin A \cdot \sin B = \cos (A - B) - \cos (A + B)

\sin (168^{\circ}) \cdot \sin (- 78^{\circ}) = - \frac{{\cos 90}^{\circ} - {\cos 246}^{\circ}}{2}

= \frac{{\cos 246}^{\circ}}{2}

= \frac{\cos (270^{\circ} - 24^{\circ})}{2}

= - \frac{p}{2}

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