trig. identity: Verify that 2sinA-cscA= sinA-cotA/cscA

[aplubin]

I need help solving a trig. identity please.

1) Verify: 2sinA-cscA= sinA-cotA/cscA

2) Verify: 2tanX/tan2X= 2-sec squaredX

Thanks in advance!

 

[skeeter]

check the terms in the first equation (see if you copied it correctly) ... it's not an identity.


2tanx/tan(2x) =

since tan(2x) = 2tanx/(1 - tan²x) ...

(2tanx)*(1 - tan²x)/(2tanx) =

1 - tan²x =

since 1 + tan²x = sec²x ...

1 - (sec²x - 1) =

2 - sec²x

 

[aplubin]

reply

Thank you so much! The first one is 2sinA - cscA = sinA - (cotA/cscA)

I would also love you forever if you helped me with this one -

cscX=-5/3 cosX is positive
cotY= 7/24 cosY is positive

find: cos(y-x) and sin(x-y)

 

[soroban]

Re: reply

Hello, aplubin!

$\displaystyle \csc x\,=\,-\frac{5}{3}\;\;\cos x$ is positive.
$\displaystyle \cot y\,=\,\frac{7}{24}\;\;\cos y$is positive.

Find: $\displaystyle \:\cos(y\,-\,x)$ and $\displaystyle \sin(x\,-\,y)$

$\displaystyle \csc x$ is negative. .$\displaystyle x$ is in quadrant 3 or 4.
$\displaystyle \cos x$ is positive. .$\displaystyle x$ is in quadrant 1 or 4.
. . Hence: $\displaystyle x$ is in quadrant 4.

Since $\displaystyle \csc x\:=\:-\frac{5}{3} \:=\:\frac{hyp}{opp}$,
. . we have: $\displaystyle \,opp\,=\,-3,\:hyp\,=\,5$ . . . and Pythagorus gives us: $\displaystyle \,adj\,=\,4$
Hence: \(\displaystyle \L\:\sin x\,=\,-\frac{3}{5},\:\cos x\,=\,\frac{4}{5}\;\) [1]


$\displaystyle \cot y$ is positive. .$\displaystyle y$ is in quadrant 1 or 3.
$\displaystyle \cos y$ is positive. .$\displaystyle y$ is in quadrant 1 or 4.
. . Hence: $\displaystyle \,y$ is in quadrant 1.

Since $\displaystyle \cot y \:=\:\frac{7}{24}\:=\:\frac{adj}{opp}$
. . we have: $\displaystyle \,opp\,=\,24,\:adj\,=\,7$ . . . and Pythagorus gives us: $\displaystyle \,hyp\,=\,25$
Hence: \(\displaystyle \L\,\sin y\,=\,\frac{24}{25},\:\cos y\,=\,\frac{7}{25}\;\) [2]


Identity: $\displaystyle \:\cos(y\,-\,x)\;=\;\cos(y)\cdot\cos(x)\,+\,\sin(x)\cdot\sin(y)$

We have: \(\displaystyle \:\cos(y\,-\,x)\;=\;\left(\frac{7}{25}\right)\left(\frac{4}{5}\right)\,+\,\left(\frac{24}{25}\right)\left(-\frac{3}{5}\right)\;=\;\frac{28}{125}\,-\,\frac{72}{125}\;=\;\fbox{-\frac{44}{125}}\)


Identity: $\displaystyle \:\sin(x\,-\,y)\;=\;\sin(x)\cdot\cos(y)\,-\,\sin(y)\cdot\cos(x)$

We have: \(\displaystyle \:\sin(x\,-\,y) \;=\;\left(-\frac{3}{5}\right)\left(\frac{7}{25}\right)\,-\,\left(\frac{24}{25}\right)\left(\frac{4}{5}\right) \;=\;-\frac{21}{125}\,-\,\frac{96}{125}\;=\;\fbox{-\frac{117}{125}}\)

 

[skeeter]

Re: reply

The first one is 2sinA - cscA = sinA - (cotA/cscA)

as I said before, this is not an identity ... let A = pi/6, the two sides of the equation are not equal.

2sin(pi/6) - csc(pi/6) = 2(1/2) - 2 = -1

sin(pi/6) - [cot(pi/6)/csc(pi/6)] = 1/2 - [sqrt(3)/2] = [1 - sqrt(3)]/2

 

[aplubin]

I'm soooo sorry! I wrote the problem slightly wrong.

The correct one is:

Verify: 2sinA - cscA = sinA - (cotA/secA)

 

[skeeter]

\(\displaystyle \L \sin{x} - \frac{\cot{x}}{\sec{x}} =\)

\(\displaystyle \L \sin{x} - \frac{\cos^2{x}}{\sin{x}} =\)

\(\displaystyle \L \frac{\sin^2{x} - \cos^2{x}}{\sin{x}} =\)

\(\displaystyle \L \frac{\sin^2{x} - (1 - \sin^2{x})}{\sin{x}} =\)

\(\displaystyle \L \frac{2\sin^2{x} - 1}{\sin{x}} =\)

\(\displaystyle \L \frac{2\sin^2{x}}{\sin{x}} - \frac{1}{\sin{x}} =\)

\(\displaystyle \L 2\sin{x} - \csc{x}\)

 

[aplubin]

Thank you!!! You really made my day =]