Question

how to solve if 3sinx+4cosx=√17 find the value of 3cosx-4sinx​

Answer 1

Answer:

Answer is 2root2

$2 \sqrt[]{2}$

Step-by-step explanation:

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Answer 2

The value of $3\cos x-4\sin x​=$ $2\sqrt{2}$

Step-by-step explanation:

We have,

$3\sin x+4\cos x=\sqrt{17}$                 ..........(1)

Let $3\cos x-4\sin x​=a$             ..........(2)

To  find,  the value of $3\cos x-4\sin x​=$ ?

Squaring and adding equations (1) and (2), we get

$(3\sin x+4\cos x)^2+(3\cos x-4\sin x​)^2=(\sqrt{17})^2+a^2$

⇒ $9\sin^2 x+16\cos^2 x+12\cos x\sin x​+9\sin^2 x+16\cos^2 x-12\cos x\sin x=17+a^2$

⇒ $9\sin^2 x+16\cos^2 x​+9\sin^2 x+16\cos^2 x=17+a^2$

⇒ $9(\sin^2 x+\cos^2 x)​+16(\sin^2 x+\cos^2 x)=17+a^2$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A=1$

$9(1)​+16(1)=17+a^2$

⇒ $25=17+a^2$

⇒ $a^2=25-17=8$

⇒ $a^2=(2\sqrt{2})^2$

⇒ a = $2\sqrt{2}$

Thus,  the value of $3\cos x-4\sin x​=$ $2\sqrt{2}$