Double Angle Proofs - Explained in Trig

[Hogart]

Hello,

I'm currently trying to convert polar equations to Cartesian equations and vise-versa. I noticed an equation needed a Trig ID I am not familiar with:

sin(2theta) = 2sin(theta)*cos(theta)

cos(2theta) = cos^2(theta) - sin^2(theta)

My apologies if that looks like an eye soar; I'm new here and don't know how to use those fancy characters. Anyway, I don't recall these ID's. I might have not been paying attention in Trig.

If anyone can explain to me how they are equal I would greatly appreciate it!

 

[BloodyFrozen]

Use the addition (or if you want subtraction) identity.

sin( a+b ) = sin(a)cos(b) + cos(a)sin(b)
cos( a+b ) = cos(a)cos(b) - sin(a)sin(b)

Setting a for theta and b also for theta. See if you can figure it out...

sin( theta+theta ) = ??
cos( theta+theta) = ??

 

[stallionx]

Hello,

I'm currently trying to convert polar equations to Cartesian equations and vise-versa. I noticed an equation needed a Trig ID I am not familiar with:

sin(2theta) = 2sin(theta)*cos(theta)

cos(2theta) = cos^2(theta) - sin^2(theta)

My apologies if that looks like an eye soar; I'm new here and don't know how to use those fancy characters. Anyway, I don't recall these ID's. I might have not been paying attention in Trig.

If anyone can explain to me how they are equal I would greatly appreciate it!

If a vector is r*Cos(A) i + r*sin(A) j

and the other is :

r*Cos(B) i + r*sin(B) j

and you dot product them

and A is bigger than B by A-B difference

r**2 cos A cos B + r**2 Sin A Sin B = r**2 cos( angle between)

cos A cos B + sin A sin B = cos(A-B) if B=-B

it is easy toı show that

Cos(A+B) = cosA cos B - Sin A sin B

And where sin**2 + cos**2 = 1

you can find

Sin(A+B)=Sin A cosB + cos A sin B

 

[stallionx]

Also Cos(90 - (A+B)) = Sin(A+B)
cos((90-A ) - B)
Apply result from proof

cos (90-A) cos(B)+sin (90-A) sin(B) = Sin(A+B)

SinA cosB+cosA sin B = Sin(A+B)

 

[CellCoree]

Hello there,

I am familiar with trigonometry and its various identities, including the double angle identities. These identities are used to simplify trigonometric expressions and equations, and they are derived from the basic trigonometric functions of sine, cosine, and tangent.

To understand the double angle identities, it is important to first understand the concept of double angles. A double angle is an angle that is twice the measure of another angle. For example, if we have an angle theta, the double angle would be 2theta.

The first identity you mentioned, sin(2theta) = 2sin(theta)*cos(theta), is known as the double angle formula for sine. It states that the sine of a double angle is equal to twice the product of the sine and cosine of the original angle. This can be derived using the sum identity for sine, which states that sin(A+B) = sin(A)cos(B) + sin(B)cos(A). If we let A = B = theta, we get sin(2theta) = 2sin(theta)*cos(theta).

Similarly, the second identity, cos(2theta) = cos^2(theta) - sin^2(theta), is known as the double angle formula for cosine. It can be derived using the difference identity for cosine, which states that cos(A-B) = cos(A)cos(B) + sin(A)sin(B). Again, if we let A = B = theta, we get cos(2theta) = cos^2(theta) - sin^2(theta).

These identities may seem complex at first, but with practice and understanding of the basic trigonometric functions, they can be easily applied in converting between polar and Cartesian equations. I hope this explanation helps you in your understanding of double angle proofs.