Question 3.2 says prove that .... =1 and we stumped

[godfreyjh]

Friends

We have tried various approaches to this question however, we are simply put, stumped.

We would appreciate if someone would give this question a go. If indeed it is possible to prove that all of this is equal to 1

The question comes from a South African Maths N4 Nated Exam Nov 2016 (Engineering Maths)

Question 3.2

Prove that \(\displaystyle \dfrac{sin 2 (\theta)}{2}+\dfrac{sin^3 (\theta) +cos^3 (\theta)}{sin (\theta) + cos (\theta)}=1\)

P.S. We busy with exam prep and this paper frustratingly does not have a memo. So we can't even peep and see what the examiner did to prove this.

Thanks in advance for your time and trouble.

As always it is appreciated.

Regards Godfrey

 

[Mrspi]

Friends

We have tried various approaches to this question however, we are simply put, stumped.

We would appreciate if someone would give this question a go. If indeed it is possible to prove that all of this is equal to 1

The question comes from a South African Maths N4 Nated Exam Nov 2016 (Engineering Maths)

Question 3.2

Prove that \(\displaystyle \dfrac{sin 2 (\theta)}{2}+\dfrac{sin^3 (\theta) +cos^3 (\theta)}{sin (\theta) + cos (\theta)}=1\)

P.S. We busy with exam prep and this paper frustratingly does not have a memo. So we can't even peep and see what the examiner did to prove this.

Thanks in advance for your time and trouble.

As always it is appreciated.

Regards Godfrey

What have you tried? Did you use the double-angle formula for sin 2x? sin 2x = 2 * (sin x cos x)

If you make that substitution in the first fraction, it simplifies nicely.

Then, you'll need to review how one factors the sum of two cubes:
a³ + b³ = (a + b) * (a² - a*b + b²)

Use this factorization to rewrite the numerator of the second fraction....and that fraction will simplify nicely, too. THEN, you'll have an addition to do, and a look at the Pythagorean identities....sin² x + cos² = 1 (this should be sin² x + cos² x = 1.......edit)

I hope this helps. If you're still having trouble, please show us what you've tried, so we can see where you may be going astray.

 

[Bob Brown MSEE]

\(\displaystyle \dfrac{sin (2 \theta)}{2}+\dfrac{sin^3 (\theta) +cos^3 (\theta)}{sin (\theta) + cos (\theta)}=1\)

 

[godfreyjh]

Thank so much

What have you tried? Did you use the double-angle formula for sin 2x? sin 2x = 2 * (sin x cos x)

If you make that substitution in the first fraction, it simplifies nicely.

Then, you'll need to review how one factors the sum of two cubes:
a³ + b³ = (a + b) * (a² - a*b + b²)

Use this factorization to rewrite the numerator of the second fraction....and that fraction will simplify nicely, too. THEN, you'll have an addition to do, and a look at the Pythagorean identities....sin² x + cos² = 1 (this should be sin² x + cos² x = 1.......edit)

I hope this helps. If you're still having trouble, please show us what you've tried, so we can see where you may be going astray.

Thank you so very much.

Your hits and suggestions put us on a path which never occurred to us when we trying to crack this one.

Thank you again for your time trouble and explanation which is immensely appreciated

Regards

From Godfrey and my Son Richard