Help needed on practice test

[mkw92]

Hello,

I am in a High School Pre-Calculus class and we were given a practice test to complete, due on Monday. However, I am struggling to understand how to solve several problems. I mostly don't know where to start...She said we could ask tutor's, friend's, family, etc. but none of my family knows how to do it and my friends are just as lost as I am! Any help given would be greatly appreciated!

Verify using Trigonometric identities: Sin/(Cot+Csc)=1-Cos

Use the half angle formulas to solve: Cos 22.5 degrees

For this one, I know the start should have Cos (45/2) since that would equal 22.5, but I'm not sure what to plug into the rest of the formula.

Use the Sum to Product formula to solve: Sin 9 x - Sin 7 x

Use the Product to Sum formula to solve: 8 Sin (pi/8) Sin (pi/8)


Thanks soooo much!

The Product to Sum formula is: Sin U Sin V= (1/2)[cos (U-V) - Cos (U+V)]
The Sum to Product formula is: Sin U - Sin V= 2 Cos [(U+V)/2] Sin [(U-V)/2]
And the Half Angle formula is: Cos (U/2)= + or - square root of [(1+Cos U)/2]

 

[wjm11]

Verify using Trigonometric identities: Sin/(Cot+Csc)=1-Cos

We have a mix of four different trig functions here. I like to start by trying to eliminate/convert some of them. Usually that means I’ll try to convert things into sine and cosine. Consider the left side denominator:

Cot+Csc = (cos/sin) + (1/sin) = (cos + 1)/sin

So our problem (left side) can be written as

Sin/(Cot+Csc) = sin/[(cos + 1)/sin] = (sin)[sin/(cos + 1)] = sin^2/(cos + 1)

sin^2/(cos + 1) = (1 – cos^2)/(cos + 1)

Now factor the top:

(1 – cos)(1 + cos)/(1 + cos) = 1 – cos Q.E.D.

 

[chrisr]

I will walk you through the first one,
so you may be able to continue through the others...

SinA/{CotA+CosecA} = 1-CosA how do we prove this ?

First, since we should "end up" with CosA,
we first write what CotA and CosecA are in terms of SinA, CosA or TanA, the obvious ones.

Therefore SinA/{[1/(tanA)] + 1/(SinA)} is SinA/([CosA/SinA] + 1/SinA)

because tanA is SinA/CosA and CotA is that inverted (upside down).

underneath the line we can add the fractions to get SinA/{(CosA+1)/SinA}.

Dividing by a fraction is the same as multiplying by a fraction turned upside down.
Divide by 2 is "multiply by" a half etc.

So this is Sin[sup:1vhl0ndy]2[/sup:1vhl0ndy]A/(1+CosA)

Now use Sin[sup:1vhl0ndy]2[/sup:1vhl0ndy]A = 1-Cos[sup:1vhl0ndy]2[/sup:1vhl0ndy]A from Cos[sup:1vhl0ndy]2[/sup:1vhl0ndy]A + Sin[sup:1vhl0ndy]2[/sup:1vhl0ndy]A = 1.
The reason we do this is because the Cos under the line must cancel with Cos to leave us without a fraction.

Therefore, we must factorise 1-Cos[sup:1vhl0ndy]2[/sup:1vhl0ndy]A, which is the difference of two squares

1-Cos[sup:1vhl0ndy]2[/sup:1vhl0ndy]A = (1-CosA)(1+CosA).

Notice that Cos[sup:1vhl0ndy]2[/sup:1vhl0ndy]A is the way (CosA)[sup:1vhl0ndy]2[/sup:1vhl0ndy] is written in maths.

Therefore (1+CosA)/(1+CosA) is 1, leaving us with 1-CosA.

That's the idea.

For Cos22.5[sup:1vhl0ndy]0[/sup:1vhl0ndy], you have already discovered the value "U" for yourself,
though there are other half or double angle formulae you could use.
That is much simpler than the example above.
Now you do your final calculation with the right hand side using that U.

For the SinU - SinV example, what are the values of "U" and "V" ?
Those are the values you use when multiplying the Cos by the Sin on the right hand side.

For the last one, are U and V the same?

 

[mkw92]

For the SinU - SinV example, what are the values of "U" and "V" ?
Those are the values you use when multiplying the Cos by the Sin on the right hand side.

The "Sin U" would be equal to 9x and the "Sin V" would be equal to 7x. So would I solve the equation leaving the variables there as U and V? ex. 2 Cos [(9x+7x)/2] Sin [(9x-7x)/2] But that doesn't seem to make sense...how does that work?

And yes, in the last one, U and V are the same.

Thanks so much! The first one (Sin/(Cot+Csc)=1-Cos) makes so much more sense now!!!

 

[chrisr]

Yes, the SinU - SinV one cannot be completed until you have a value for x.
You see, the U and V values themselves need x,
so for those you must have x.
So that's all you need to do if you do not have x, place U and V exactly as you have done.

Once you have that value, you can then complete the calculations for Cos(angle) and Sin(angle)
using the calculator.

To be honest, you can calculate the answers without having to use the identities,
at least for the examples you have there,
but it is practice, and it shows that you get the same overall answers.

In the cases where you have actual values for U and V, you can now perform the calculations,
if you like to, you can work out both sides of the equations and see if your answers match.

 

[chrisr]

One thing I forgot to mention, pi is 180 degrees,
I'm not sure how much experience you have using the two sets of units for measuring angles,
degrees and radians. 360 degrees = 2(pi) radians.

pi/8 is also 22.5 degrees.