Question

4cos^3 - 3sin50 pls answer this question

Answer 1

Answer:Sin 50°=Sin(90-40)=Cos 40°

4 Cos^3(40)-3 Cos 40= Cos3(40)=

Cos 120°=-1/2

Answer 2

Correct Question: Find the value of $4cos^3 40^{0}-3sin50^0$.

Answer:

The value of the trigonometric function $4cos^3 40^{0}-3sin50^0$ is $-\frac{1}{2}$.

Step-by-step explanation:

Consider the trigonometric function as follows:

$4cos^3 40^{0}-3sin50^0$ ...... (1)

We know that $cos(90^0- x)=sinx$

Then equation (1) can be written as,

$4cos^3 40^{0}-3sin50^0 = 4cos^3 40^{0}-3cos(90-50)$

                              $= 4cos^3 40^{0}-3cos40^0$ ...... (2)

Now, using the identity, $cos3x=4cos^3x-3cosx$, i.e.,

⇒ $4cos^3x=cos3x+3cosx$

Substitute the value $cos3x+3cosx$ for $4cos^3x$ in the equation (2) as follows:

⇒ $4cos^3 40^{0}-3sin50^0= cos3 (40)^0+3cos40^0-3cos40^0$

Simplify as follows:

⇒ $4cos^3 40^{0}-3sin50^0= cos120^0$

⇒ $4cos^3 40^{0}-3sin50^0= -\frac{1}{2}$ (Since $cos120^0=-\frac{1}{2}$)

Therefore, the value of  the trigonometric function $4cos^3 40^{0}-3sin50^0$ is $-\frac{1}{2}$.

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