Question

If sinA+cosA = /2 cos A Then wt is SinA - cos A

Answer 1

Answer:

If cosA+sinA= √2cosA, then what is cosA-sinA?

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If you want answer in terms of trigonometric function,

cosA+sinA=2–√cosA

Squaring,

cos2A+sin2A+2cosAsinA=2cos2A

Using cos2A=1−sin2A

1−sin2A+sin2A+2cosAsinA=2(1−sin2A)

1+2cosAsinA=2−2sin2A

2cosAsinA=1−2sin2A

1−2cosAsinA=2sin2A

cos2A+sin2A−2cosAsinA=2sin2A

(cosA−sinA)2=2sin2A

cosA−sinA=±2–√sinA

If you want answer in terms of the exact values,

cosA+sinA=2–√cosA

cosAcosA+sinAcosA=2–√

1+tanA=2–√

tanA=2–√−1

A=tan−1(2–√−1)=22.5°orπ8 radians

So,

cosA−sinA=cos22.5°−sin22.5°≈0.9238795325−0.3826834324≈0.5411961001 Please mark the answer as the branliest

Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given that
• $\sin\theta + \cos\theta = \sqrt{2}\cos\theta$

To Find:

• We have to find the value of
• $\sin\theta -\cos\theta$

Solution:

We have been given that

$\sin\theta + \cos\theta = \sqrt{2}\cos\theta$

$\sin\theta = \sqrt{2}\cos\theta - \cos\theta$

$\sin\theta = \cos\theta(\sqrt{2} - 1)$

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Multiplying by ( $\sqrt{2} + 1$) on Both sides

$\sin\theta(\sqrt{2} + 1) = \cos\theta(\sqrt{2} -1)(\sqrt{2} +1)$

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Using [ ( a - b )( a + b ) = a² - b² ]

$\sin\theta(\sqrt{2}+1) = \cos\theta[\: {(\sqrt{2})}^{2} - {(1)}^{2}\:]$

$\sin\theta(\sqrt{2} +1) = \cos\theta(2-1)$

$\sqrt{2}\sin\theta + \sin\theta = \cos\theta$

$\cos\theta - \sin\theta = \sqrt{2}\sin\theta$

$\sin\theta - \cos\theta = ( - \sqrt{2}\sin\theta)$

Hence $\sin\theta - \cos\theta = ( - \sqrt{2}\sin\theta)$

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$\boxed{</p><p></p><p>\begin{minipage}{6 cm}</p><p></p><p>Fundamental Trigonometric Identities \\ \\</p><p></p><p> \sin^2\theta + \cos^2\theta=1 \\ \\</p><p></p><p>1+\tan^2\theta = \sec^2\theta \\ \\</p><p></p><p>1+\cot^2\theta = \text{cosec}^2 \, \theta </p><p></p><p>\end{minipage}</p><p>}$

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