Can you draw the
The solution for this:
Establishing identities
1.) Sin²∅ (1+cot²∅) = 1
2.) (tan²B+1) cos²B = 1
3.) tan x
---------- = sin x
sec x
4.) 1 1
------- + ------- = 2csc²∅
1+cos∅ 1-cos
Please help me, i cant understand it.
Most of these appear to be based on the Pythagorean identity
sin^2 x + cos^2 x = 1 , and its variations
e.g. if I divide the above by sin^2 x , I get
1 + cot^2 x = csc^2 x
and that is what I see in the LS of #1
Use LS for left side and RS for right side, work each side independently . I usually start with the more complicated looking side and try to change it to the expression on the other side.
Sometimes you have to work on both sides.
As long as you reach some point where both sides are the same, you have done your job.
so #1
LS = sin^2 Ø (1+cot^2 Ø)
= sin^2 Ø (csc^2 Ø)
= 1
= RS
#2, just one of the other variations, you try it
#3. just based on basic definitions,
LS = (sinx/cosx) / (1/cosx)
= (sinx/cosx)(cosx/1)
= sinx
= RS
#4.
LS = 1/(1+cosx) + 1/(1-cosx)
form a common denominator and simplify
= (1 - cosx + 1 + cosx)/(sin^2 x)
= 2/sin^2 x
= 2csc^2 x
= RS
One more clue.
Since all 6 trig functions can be defined in terms of sin and cos, I often change everything to sines and cosines, unless I see an obvious definition such as
1 + cot^2 Ø = csc^2 Ø
in #1
How it looks like , ? #1 to get this 1 + cot^2 x = csc^2 x ??
I get it!
Sure, I can help you with these identities. Let's go through each of them step by step:
1.) Sin²∅ (1+cot²∅) = 1
To prove this identity, we can rewrite cot²∅ as (cos²∅ / sin²∅). So, the equation becomes:
Sin²∅ (1+cos²∅ / sin²∅) = 1
Now, we can simplify by multiplying sin²∅ with (sin²∅ + cos²∅):
sin²∅sin²∅ + sin²∅cos²∅ / sin²∅ = 1
sin⁴∅ + cos²∅ = 1 (using the property sin²∅ + cos²∅ = 1)
So, the identity Sin²∅ (1+cot²∅) = 1 is proven.
2.) (tan²B+1) cos²B = 1
To solve this identity, let's simplify the left side of the equation:
(tan²B+1) cos²B = (sin²B/cos²B + 1) cos²B
Simplifying further, we get:
(sin²B + cos²B) = 1 (using the Pythagorean identity sin²B + cos²B = 1)
So, the identity (tan²B+1) cos²B = 1 has also been proven.
3.) tan x / sec x = sin x
To solve this identity, let's rewrite tan x as sin x / cos x and sec x as 1 / cos x:
(sin x / cos x) / (1 / cos x) = sin x
Now, we can simplify:
(sin x / cos x) * (cos x / 1) = sin x
sin x = sin x
So, the identity tan x / sec x = sin x is proven.
4.) 1 / (1+cos∅) + 1 / (1-cos∅) = 2csc²∅
To prove this identity, let's find a common denominator and simplify:
[(1 - cos∅) + (1 + cos∅)] / [(1 + cos∅)(1 - cos∅)] = 2csc²∅
[2] / [1 - cos²∅] = 2csc²∅ (using the Pythagorean identity sin²∅ = 1 - cos²∅)
2 / sin²∅ = 2csc²∅ (using the reciprocal identity csc∅ = 1 / sin∅)
1 / sin²∅ = csc²∅
Since the left side of the equation is the reciprocal of the right side, the identity 1 / (1+cos∅) + 1 / (1-cos∅) = 2csc²∅ is proven.
I hope this helps you understand and solve these identity problems!