If sin theta cos theta cos theta sin theta = 2

The Trigonometric Identities are equations that are true for Right Angled Triangles. (If it is not a Right Angled Triangle go to the Triangle Identities page.)

Each side of a right triangle has a name:

Adjacent is always next to the angle

And Opposite is opposite the angle

We are soon going to be playing with all sorts of functions, but remember it all comes back to that simple triangle with:

• Angle θ
• Hypotenuse
• Adjacent
• Opposite

Sine, Cosine and Tangent

The three main functions in trigonometry are Sine, Cosine and Tangent.

They are just the length of one side divided by another

For a right triangle with an angle θ :

For a given angle θ each ratio stays the same
no matter how big or small the triangle is

When we divide Sine by Cosine we get:

sin(θ)cos(θ) = Opposite/HypotenuseAdjacent/Hypotenuse = OppositeAdjacent = tan(θ)

So we can say:

That is our first Trigonometric Identity.

Cosecant, Secant and Cotangent

We can also divide "the other way around" (such as Adjacent/Opposite instead of Opposite/Adjacent):

sin(θ) = 2/4, and csc(θ) = 4/2

Because of all that we can say:

sin(θ) = 1/csc(θ)

cos(θ) = 1/sec(θ)

tan(θ) = 1/cot(θ)

And the other way around:

csc(θ) = 1/sin(θ)

sec(θ) = 1/cos(θ)

cot(θ) = 1/tan(θ)

And we also have:

Pythagoras Theorem

For the next trigonometric identities we start with Pythagoras' Theorem:

The Pythagorean Theorem says that, in a right triangle, the square of a plus the square of b is equal to the square of c: a2 + b2 = c2

Dividing through by c2 gives

a2 c2 + b2 c2 = c2 c2

This can be simplified to:

( a c )2 + ( b c )2 = 1

Now, a/c is Opposite / Hypotenuse, which is sin(θ)

And b/c is Adjacent / Hypotenuse, which is cos(θ)

So (a/c)2 + (b/c)2 = 1 can also be written:

Note:

• sin2 θ means to find the sine of θ, then square the result, and
• sin θ2 means to square θ, then do the sine function

Using 4 decimal places only:

• sin(32°) = 0.5299...
• cos(32°) = 0.8480...

Now let's calculate sin2 θ + cos2 θ:

0.52992 + 0.84802 = 0.2808... + 0.7191...

= 0.9999...

We get very close to 1 using only 4 decimal places. Try it on your calculator, you might get better results!

Related identities include:

sin2 θ = 1 − cos2 θ
cos2 θ = 1 − sin2 θ
tan2 θ + 1 = sec2 θ
tan2 θ = sec2 θ − 1
cot2 θ + 1 = csc2 θ
cot2 θ = csc2 θ − 1

The identities mentioned so far can be remembered using one clever diagram called the Magic Hexagon:

But Wait ... There is More!

There are many more identities ... here are some of the more useful ones:

Opposite Angle Identities

sin(−θ) = −sin(θ)

cos(−θ) = cos(θ)

tan(−θ) = −tan(θ)

Double Angle Identities

Half Angle Identities

Note that "±" means it may be either one, depending on the value of θ/2

Angle Sum and Difference Identities

Note that

sin(A B) = sin(A)cos(B) cos(A)sin(B)

cos(A B) = cos(A)cos(B) sin(A)sin(B)

tan(A B) = tan(A) tan(B)1 tan(A)tan(B)

cot(A B) = cot(A)cot(B) 1cot(B) cot(A)

Triangle Identities

There are also Triangle Identities which apply to all triangles (not just Right Angled Triangles)

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General discussion to determine the sign of #=>sintheta+costheta#

From the given condition

#sintheta+sqrt2sintheta=costheta#

#=>(sqrt2+1)sintheta=costheta#

#:.sintheta/costheta=1/(sqrt2+1)....(1)#

#tantheta=1/(sqrt2+1)>0#

#tantheta# being positve #theta# should be in 1st or 3rd quadrant.when #theta# is in the 1st quadrant is positive but in 3rd quadrant #sintheta and costheta# both negative and the the value #sintheta+costheta#should be negative.

So # color(red)(sintheta+costheta" will be either +ve or -ve")#

Method -I (A tricky approach)

From equation (1) have

#sintheta/costheta=1/(sqrt2+1)....(1)#

and

Inverting and ratiolising numerator of RHS

#costheta/sintheta=sqrt2+1=1/(sqrt2-1)....(2)#

Adding equation(1) and equation (2)we get

#sintheta/costheta+costheta/sintheta=1/(sqrt2+1)+1/(sqrt2-1)#

#(sin^2theta+cos^2theta)/(sinthetacostheta)=(2sqrt2)/((sqrt2-1)(sqrt2+1))=(2sqrt2)/1#

Inverting and rearranging we get

#(2sinthetacostheta)/(sin^2theta+cos^2theta)=1/(sqrt2)#

Adding 1 on both sides we get

#1+(2sinthetacostheta)/(sin^2theta+cos^2theta)=1+1/(sqrt2)#

#(sintheta+costheta)^2/(sin^2theta+cos^2theta)=(sqrt2+1)/sqrt2#

#=>(sintheta+costheta)^2=(sqrt2+1)/sqrt2#

#=>sintheta+costheta=+-sqrt((sqrt2+1)/sqrt2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method - II

Given
#sintheta-costheta=-sqrt2sintheta#

Dividing both sides by# sqrt2 " "#we have

#=>1/sqrt2sintheta-1/sqrt2costheta=-sintheta#

(Considering unit of angle in degree.)

#=>cos(45)sintheta-sin(45)costheta=-sintheta...(1)#

#=>sin(theta-45)=sin(-theta)#

#=>theta-45=-theta#

#=>2theta=45#

#=>theta=22.5#

Another solution is posible satisfying positive value of #tantheta# when #theta# is in third quadrant

Then from eq (1)

#sin(theta-45)=sin(360-theta)#
#=>theta=405/2=180+22.5#

Now

#sintheta+costheta#

#=sqrt2(1/sqrt2sintheta+1/sqrt2costheta)#
#=sqrt2sin(theta+45)....(2)#

#color (blue)("when " " theta=22.5#

Inserting #theta=22.5#

#=sqrt2sin(22.5+45)#

#=sqrt2sin(90-22.5)#

#=sqrt2cos(22.5)#

#=sqrt2sqrt(1/2(1+cos45))#

#=sqrt(1+cos45)#

#=sqrt(1+1/sqrt2)#

#=sqrt((sqrt2+1)/sqrt2)#

#color (green)("Again when " " theta=180+22.5#

we put #theta=180+22.5#

in eq(2)

we get

#=>sintheta+costheta =sqrt2sin(theta+45)#

#=sqrt2sin(180+22.5+45)#

#=sqrt2sin(180+22.5+45+22.5-22.5)#

#=sqrt2sin(270-22.5)#

#=-sqrt2cos(22.5)#

#=-sqrt2(sqrt(1/2(1+cos45)))#

#=-sqrt2(sqrt(1/2(1+1/sqrt2)))#

#=-sqrt(1+1/sqrt2)#

#=-sqrt((sqrt2+1)/sqrt2)#

So combining these two we get

#sintheta+costheta=+-sqrt((sqrt2+1)/sqrt2)#