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Can anyone help me with these questions?

I can do 1 part a). Its the rest im having trouble with since i havent studied this type of trig.

1 a) prove the identity cot (1/2A)- tan (1/2A)=2cotA

b) By choosing a suitable numerical value for A, show that tan 15 is a root of the quadratic equation t^2 + 2√3t -1 = 0

2 a) By using the substitution t = tan (1/2x), prove that cosec x - cot x= tan (1/2x)

b) Use the result to show that tan 15= 2- √3

I think its to do with the double angle formula but i havent studied this at all. Thanks. Please help asap.

I can do 1 part a). Its the rest im having trouble with since i havent studied this type of trig.

1 a) prove the identity cot (1/2A)- tan (1/2A)=2cotA

b) By choosing a suitable numerical value for A, show that tan 15 is a root of the quadratic equation t^2 + 2√3t -1 = 0

2 a) By using the substitution t = tan (1/2x), prove that cosec x - cot x= tan (1/2x)

b) Use the result to show that tan 15= 2- √3

I think its to do with the double angle formula but i havent studied this at all. Thanks. Please help asap.

bluebird

1 a) prove the identity cot (1/2A)- tan (1/2A)=2cotA

.

1 a) prove the identity cot (1/2A)- tan (1/2A)=2cotA

.

cot2θ can be written as (1-tan²θ )/2tanθ

letting 2θ=A

cotA = (1-tan²(A/2))/2tan(A/2)

dividing out:

1/(2tan(A/2)) -tan²(A/2)/(2tan(A/2))

½cot(A/2) - ½tan(A/2) = cot A

multiply by 2:

cot(A/2) -tan(A/2) = 2cotA

b) let A = 30° let t = tan15°

1/t - t = 2cot30°

1/t - t = 2√3

1 - t² = 2√3t

t² + 2√3t -1 = 0

2a) to prove that cosec x - cot x= tan (1/2x):

the LHS can be written as 1/sinx - cosx/sinx

or (1- cosx)/sinx

cosx can be written as 1 - 2sin²(x/2)

so 1 - cosx can be written as 2sin²(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/sinx

now sinx can be written as 2sin(x/2)cos(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/2sin(x/2)cos(x/2)

which simplifies to sin(x/2)/cos(x/2) which is tan(x/2)

b) Use the result to show that tan 15= 2- √3

if x/2 is 15° then x will be 30°...cot30° is √3 and cosec30° is 2....

Original post by the bear

cot2θ can be written as (1-tan²θ )/2tanθ

letting 2θ=A

cotA = (1-tan²(A/2))/2tan(A/2)

dividing out:

1/(2tan(A/2)) -tan²(A/2)/(2tan(A/2))

½cot(A/2) - ½tan(A/2) = cot A

multiply by 2:

cot(A/2) -tan(A/2) = 2cotA

b) let A = 30° let t = tan15°

1/t - t = 2cot30°

1/t - t = 2√3

1 - t² = 2√3t

t² + 2√3t -1 = 0

2a) to prove that cosec x - cot x= tan (1/2x):

the LHS can be written as 1/sinx - cosx/sinx

or (1- cosx)/sinx

cosx can be written as 1 - 2sin²(x/2)

so 1 - cosx can be written as 2sin²(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/sinx

now sinx can be written as 2sin(x/2)cos(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/2sin(x/2)cos(x/2)

which simplifies to sin(x/2)/cos(x/2) which is tan(x/2)

b) Use the result to show that tan 15= 2- √3

if x/2 is 15° then x will be 30°...cot30° is √3 and cosec30° is 2....

letting 2θ=A

cotA = (1-tan²(A/2))/2tan(A/2)

dividing out:

1/(2tan(A/2)) -tan²(A/2)/(2tan(A/2))

½cot(A/2) - ½tan(A/2) = cot A

multiply by 2:

cot(A/2) -tan(A/2) = 2cotA

b) let A = 30° let t = tan15°

1/t - t = 2cot30°

1/t - t = 2√3

1 - t² = 2√3t

t² + 2√3t -1 = 0

2a) to prove that cosec x - cot x= tan (1/2x):

the LHS can be written as 1/sinx - cosx/sinx

or (1- cosx)/sinx

cosx can be written as 1 - 2sin²(x/2)

so 1 - cosx can be written as 2sin²(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/sinx

now sinx can be written as 2sin(x/2)cos(x/2)

so the LHS is now 2sin(x/2)sin(x/2)/2sin(x/2)cos(x/2)

which simplifies to sin(x/2)/cos(x/2) which is tan(x/2)

b) Use the result to show that tan 15= 2- √3

if x/2 is 15° then x will be 30°...cot30° is √3 and cosec30° is 2....

Why did you let A= 30 degrees?

Original post by Chots

Why did you let A= 30 degrees?

because then A/2 will be 15° as required

Original post by the bear

because then A/2 will be 15° as required

Ohh, of course, thank you!

Original post by the bear

because then A/2 will be 15° as required

Edit: also - PRSOM... {sigh}

Original post by DFranklin

Did you seriously just reply to a response to a post you made 9 years ago? {Impressed}!

Edit: also - PRSOM... {sigh}

Edit: also - PRSOM... {sigh}

time flies when you are having fun

i am embarrassed about giving full solutions in those crazy days

Original post by the bear

i am embarrassed about giving full solutions in those crazy days

Although I confess that my initial thought reading this thread was "Hey, full solutions are against the rules...", and then I noticed the date.

Original post by DFranklin

Hey, I did it too in the early days.

Although I confess that my initial thought reading this thread was "Hey, full solutions are against the rules...", and then I noticed the date.

Although I confess that my initial thought reading this thread was "Hey, full solutions are against the rules...", and then I noticed the date.

hehehe exactly what i thought

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