Prove the identity
sin(x+y+z)+sin(x+y-z)+sin(x-y+z)+ sin(x-y-z) = 4 sin(x)cos(y)cos(z)
This identity is so long and after i tried to expand the left side and it just looked something crap
Thanks for you help :)
LS
= sin((x+y)+z)+sin((x+y)-z)+sin((x-y)+z)+ sin((x-y)-z)
= sin(x+y)cosz + cos(x+y)sinz
+ sin(x+y)cosz - cos(x+y)sinz
+ sin(x-y)cosz + cos(x-y)sinz
+ sin(x-y)cosz - cos(x-y)sinz
= 2sin(x+y)cosz + 2sin(x-y)cosz
= 2[cosz(sinxcosy + cosxsiny)] + 2[cosz(sinxcosy - cosxsiny)]
= 2sinxcosycosz + 2sinycosxcosz + 2sinxcosycosz - 2 sinycosxcosz
= 4 sinxcosycosz
= RS
I once had a great geometry teacher back in high school in Seattle. He was old and close to retirement. He used to say, to a mostly bored and unappreciative class: "Try to see the beauty in it." A few of us did. It was my favorite subject. Trig was next.
Reiny's proof reminds me of that beauty.
Well, i tried to see the beauty in trigonometry but to me, it is just too HARD!!!!!!!!
There r so many formulae in trigonometry and how to i know which one to use.
After i stay in the desk for 15 mins, i just wanna throw this stupid book away. I cant gain anything even i try so hard.
Thanks for the comment.
Back in the days when we still wrote on a blackboards with chalk and we still had care-takers that would clean those boards at the end of the day....
We once did a proof of "in any quadrilateral the largest area is obtained when opposite angles are supplementary" and it filled about 3 sections of blackboard.
After it was done, the students drew a large picture frame around it.
Next day it was still there with a note from the care-taker.
"I looked at this, did not understand anything about it, but it sure looks like a piece of art"
Your comment reminded me of that day, thanks.
A ladder that is 6 meters long is placed against a wall. It makes an angle 0f 34 degrees with the wall. Find how high up the wall it reaches and the distnce the base of the ladder is away from the wall
To prove the given identity, we can start by using the trigonometric identity known as the sum-to-product formula. This formula states that:
sin(a) + sin(b) = 2sin((a+b)/2)cos((a-b)/2)
Applying this formula to the first two terms on the left side of the given identity, we can rewrite them as:
sin(x+y+z) + sin(x+y-z) = 2sin((x+y+z+x+y-z)/2)cos((x+y+z-(x+y-z))/2)
= 2sin((2x+2y)/2)cos((2z)/2)
= 2sin(x+y)cos(z)
Similarly, applying the sum-to-product formula to the last two terms, we can rewrite them as:
sin(x-y+z) + sin(x-y-z) = 2sin((x-y+z+x-y-z)/2)cos((x-y+z-(x-y-z))/2)
= 2sin((2x-2y)/2)cos((2z)/2)
= 2sin(x-y)cos(z)
Now, let's substitute these results back into our original equation:
2sin(x+y)cos(z) + 2sin(x-y)cos(z) = 4sin(x)cos(y)cos(z)
Both sides of the equation match, which proves the given identity:
sin(x+y+z) + sin(x+y-z) + sin(x-y+z) + sin(x-y-z) = 4sin(x)cos(y)cos(z)
Therefore, the identity is proven.