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£10/h
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Tutor profile verified
Response Time 24h
Lessons offered by Jitendra
• Individual
The lessons will be held
Taught subjects
• Trigonometry
Levels
• Primary
• Secondary
• GCSE
• AS Level
• A Level
• Masters
• Diploma
• Doctorate

# Can teach trigonometry and you will love it... All basics will be covered

## Methodology

We will solve a question

1)If Sin2x= 3/4, then sin^3x + Cos^3x=?

Let's solve it
We have,
Sin^3x+cos^3x=(sinx+ cosx)(sin^2x+cos^2x-sinx cosx)
[This is just a^3 +b^3= (a+b) (a^2+b^2-ab)]✓

Now multiply and divide by 2 to sinx cosx
= (Sinx+cosx)(Sin^2x+cos^2x-2sinxcosx/2)
As we know
2 sinx cosx= sin2x, so
= ( sinx + cosx )(1- sin2x/2),
As we know that,

Sin^2x+ cos^2x=1
After that
=√{(1+Sin2x)}(1-sin2x/2)
Now this √{(1+Sin2x)} has come in such a way
[(Sinx+ cosx)^2= sin^2x + cos^2x+2 sinx Cosx]✓
[We know that Sin^2x+ cos^2x=1]✓
[Sin2x= 2 sinx cosx]✓
(Sinx + cosx)^2= [1+Sin2x]
(Sinx+cosx)=√(1+ Sin2x)
= √1 + 3/4(1-3/8)
As Sin2x=( 3/4) given
=5√7/16

So I will tell you how to break the question by using concepts and you will get the ans. In just few steps

2) What is the min. and max. Value of
6sinx cosx+ 4 cos 2x respectively?

Let's solve it
6 sinx cos x + 4 cos 2x = 3 sin2x+4 cos2x
as Sin 2x= 2 sinx cosx, So
6 sinx cosx= 3 sin2x
Now the TRIGNOMETRICAL EXPRESSION has been of the form

a sin2x+ cos 2x, so...

Now just see what I have written and read it fully, you will understand everything as you will finish reading the entire solution

-√{(3)^2+(4)^2}

## Rates

Rate for online lessons : £10/h
Lessons offered by Jitendra
Individual
The lessons will be held
at his home
Taught subjects
• Trigonometry
Levels
• Primary
• Secondary
• GCSE
• AS Level
• A Level
• Masters
• Diploma
• Doctorate

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