Verify the Identity or show why its not an identity

[Princezz3286]

sin theta + cos theta = (tan theta +1)/sec theta
i started with the side with the tangent....
= tan+1/(1/cos).... simplify this to get rid of the 1/cos by mult by the recrip.... this is where i get lost. Manipulation equations is not my thing....

I also have show the equation is not an identity by finding a value of x and y that both sides are defined but not equal.
I dont even know where to start with this one. What are they asking me here?
sin(x+y)=sin x + sin Y why is this not equal? isnt that how it would work with the dist prop.? im confused.... please help me

 

[galactus]

Firstly, \(\displaystyle sin(x+y)\neq sin(x)+sin(y)\)

\(\displaystyle sin(x+y)=sin(x)cos(y)+cos(x)sin(y)\)

Now, for the identity:

\(\displaystyle sin{\theta}+cos{\theta}=\frac{tan{\theta}+1}{sec{\theta}}\)

Start with the right side and show it is the left side:

\(\displaystyle \frac{tan{\theta}}{sec{\theta}}+\frac{1}{sec{\theta}}\)

\(\displaystyle tan{\theta}cos{\theta}+cos{\theta}\)

\(\displaystyle \frac{sin{\theta}}{cos{\theta}}\cdot cos{\theta}+cos{\theta}\)

\(\displaystyle sin{\theta}+cos{\theta}\)

 

[BigGlenntheHeavy]

\(\displaystyle 1) \ sin(\theta)+cos(\theta) \ = \ \frac{tan(\theta)+1}{sec(\theta)} \ (RHS) \ = \ \frac{sin(\theta)/\cos(\theta)+1}{1/cos(\theta)} \ = \ \frac{sin(\theta)+cos(\theta)}{cos(\theta)}*\frac{cos(\theta)}{1}\)

\(\displaystyle = \ sin(\theta)+cos(\theta).\)


\(\displaystyle 2) \ sin(x+y) \ = \ sin(x)cos(y)+cos(x)sin(y) \ \ne \ sin(x)+sin(y).\)

\(\displaystyle Will \ prove \ by \ Euler's \ famous \ formula, \ to \ wit: \ e^{i\theta} \ = \ cos(\theta)+isin(\theta).\)

\(\displaystyle Ergo, \ e^{i(x+y)} \ = \ (e^{ix})(e^{iy}) \ = \ [cos(x)+isin(x)][cos(y)+isin(y)]\)

\(\displaystyle = \ cos(x)cos(y)+icos(x)sin(y)+isin(x)cos(y)+i^2sin(x)sin(y),\)

\(\displaystyle = \ cos(x)cos(y)-sin(x)sin(y)+i[cos(x)sin(y)+sin(x)cos(y)].\)

\(\displaystyle But \ it \ is \ also \ true \ that \ e^{i(x+y)} \ = \ cos(x+y)+isin(x+y)\)

\(\displaystyle Two \ complex \ numbers \ are \ equal \ only \ if \ their \ real \ and \ imaginary \ parts \ are \ equal.\)

\(\displaystyle Therefore \ cos(x+y) \ = \ cos(x)cos(y)-sin(x)sin(y), \ and \ (two \ for \ a \ nickel)\)

\(\displaystyle sin(x+y) \ = \ cos(x)sin(y)+sin(x)cos(y) \ = \ sin(x)cos(y)+cos(x)sin(y). \ QED\)

\(\displaystyle Now, \ what \ equals \ cos(x-y) \ and \ sin(x-y)?\)

 

[Ted]

Relax, guys. Let's keep the discussion related to the problem at hand. If you don't like one poster's explanation, feel free to submit your own. Alternatively, if you are going to be obnoxious and unhelpful, just don't post anything at all. Thank you.