How do I find #sintheta# = #-cos^2theta# -1 in radians from #[0, 2pi]#?

How do I find #sintheta# = #-cos^2theta# -1 in radians from #[0, 2pi]#?

3 Answers
Ananda Dasgupta
Apr 19, 2018

#{3pi}/2#

Explanation:

#sin theta = -cos^2 theta-1 = -(1-sin^2 theta)-1 = sin^2theta-2 implies#

#sin^2theta-sintheta-2=0 implies (sin theta-2)(sin theta +1)=0#

Since #-1 <= sin theta <= +1#, the factor #sin theta +2# can not be zero. Thus

#sin theta +1 = 0 implies theta = sin^-1(-1) = {3pi}/2#

(note that #{3pi}/2# is the only value of #theta# in #[0,2pi]# that satisfies #sin theta = -1#)

VNVDVI
Apr 19, 2018

#theta=(3pi)/2#

Explanation:

We can get this equation in terms of one trigonometric function using trig identities. In this case, we know the identity

#sin^2theta+cos^2theta=1#

#-cos^2theta=sin^2theta-1#

So, we apply it to the right side:

#sintheta=sin^2theta-1-1#

Move everything to the right. This gives us the following:

#0=sin^2theta-sintheta-2#

Or,

#sin^2theta-sintheta-2=0#

This strongly resembles a quadratic equation; however, instead of #ax^2+bx+c#, we observe the form #asin^2theta+bsintheta+c#.

As a result, we can factor this just as we would factor a quadratic:

#(sintheta-2)(sintheta+1)=0#

We then solve the following:
#sintheta-2=0#
#sintheta+1=0#

For #sintheta-2=0:#

#sintheta=2#

Tells us this one has no solutions, as #-1<=sintheta<=1# for all #theta.#

#sintheta+1=0#

#sintheta=-1#

Holds true for #theta=(3pi)/2# in the interval #[0, 2pi]#

Nine E.
Apr 19, 2018

#Theta# = #(3pi)/2#

Explanation:

We can use one of the Pythagorean identities to write our equation in terms of a single trigonometric function. In particular, we can use:

#sin^2theta+cos^2theta = 1#, rewritten as #cos^2theta = 1-sin^2theta#.

We get:

#sintheta = -cos^2theta -1#

#sintheta = -(1-sin^2theta)-1#

#sintheta = -1+sin^2theta-1#

#sintheta = sin^2theta-2#

We'll solve as follows.

#-sin^2theta+sintheta+2 = 0#

#sin^2theta-sintheta-2=0# <--Multiplying Both Sides By Negative One

#(sintheta+1)(sintheta-2) = 0# <--Factoring the left-hand side

The product of the factors #sintheta + 1 and sintheta - 2 # is zero.
So if the equation has a solution, at least one of the factors must be zero.

#sintheta+1 = 0#
#sintheta = -1#

or

#sintheta-2 = 0#
#sintheta = 2 #

Sine has the value #-1# at #(3pi)/2#.
It NEVER has the value #2#.

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