Question

4. Simplify the expression. (6 points)

Answer 1

Answer with explanation:

Ques 4)

 We are given a trignometric expression by:

     $\dfrac{\sin^2x-1}{\cos (-x)}$

which could also be written as:

$=\dfrac{-(1-\sin^2x)}{\cos x}$

since, we know that:

$\cos (-x)=\cos x$

Also,

$1-\sin^2 x=\cos^2 x$

Hence, we get:

$\dfrac{\sin^2x-1}{\cos (-x)}=\dfrac{-\cos^2x}{\cos x}\\\\\\\dfrac{\sin^2x-1}{\cos (-x)}=-\cos x$

            The correct option is:

                        D)   -cos x        

Ques 5)

We are asked to find the solution in the interval [0,2π) of the expression:

        $\sin 2x+\sin x=0$

This expression could also be written as:

$2\sin x\cos x+\sin x=0\\\\i.e.\\\\\sin x(2\cos x+1)=0\\\\i.e.\\\\Either\ \sin x=0\ or\ 2\cos x+1=0$

If

$\sin x=0$

Then the possible values of x are:

$x=0,\pi$

and if

$2\cos x+1=0\\\\i.e.\\\\\cosx=\dfrac{-1}{2}$

We know that the cosine function is negative in second and third quadrant and the possible values where x is negative is:

$x=\dfrac{2\pi}{3}\ ,\ x=\dfrac{4\pi}{3}$

Hence, all the solutions of the given expression that will lie in the given region is:

    $x=0,\ \pi\ ,\ \dfrac{2\pi}{3}\ ,\ \dfrac{4\pi}{3}$

Answer 2

4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x

(1−sin2(x))/(sin(x)−csc(x))

sin2x+cos2x=1 1−sin2x=cos2x

cos2(x)/(sin(x)−csc(x)) csc(x)=1/sin(x) cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x)) sin2(x)− 1=-cos2(x) cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)
the answer is the letter a) -sin x

5. Find all solutions in the interval [0, 2π). (6 points)
sin2x + sin x = 0
 using a graphical tool  
the solutions   x1=0 x2=pi x3=3pi/2
the answer is the letter D) x = 0, π, three pi divided by two