Prove: sin^212+sin^221+sin^239+sin^248=1+sin^29+sin^218
Not sure what sin^212 means.
sin^2(12°)?
sin(212°)?
...
1st one
To prove the given statement, we will simplify both sides of the equation separately and show that they are equal.
Let's start by simplifying the left-hand side (LHS) of the equation:
sin^212 + sin^221 + sin^239 + sin^248
Using the trigonometric identity sin^2θ = 1 - cos^2θ, we can rewrite each term:
1 - cos^212 + 1 - cos^221 + 1 - cos^239 + 1 - cos^248
Now, let's simplify each of the terms:
1 - cos^212 can be written as sin^2(90° - 12°), and using the complementary angle identity, sin^2θ = cos^2(90° - θ), so this term becomes cos^2(12°).
Similarly, 1 - cos^221 can be rewritten as cos^2(90° - 21°) = cos^2(69°).
1 - cos^239 can be written as cos^2(90° - 39°) = cos^2(51°).
And finally, 1 - cos^248 can be rewritten as cos^2(90° - 48°) = cos^2(42°).
Now that we have simplified the left-hand side, let's simplify the right-hand side (RHS) of the equation:
1 + sin^29 + sin^218
Using the identity sin^2θ = 1 - cos^2θ, we can rewrite each term:
1 + 1 - cos^29 + 1 - cos^218
Simplifying further:
1 + 1 - cos^29 can be rewritten as 1 + sin^2(90° - 9°), which is sin^2(81°).
1 - cos^218 can be rewritten as sin^2(90° - 18°) = sin^2(72°).
Now, our equation becomes:
cos²12° + cos²21° + cos²39° + cos²48° = 1 + sin²9° + sin²18°.
Since cos²θ + sin²θ = 1 (due to the Pythagorean identity), we can substitute this identity into our equation:
1 + sin²12° + 1 + sin²21° + 1 + sin²39° + 1 + sin²48° = 1 + sin²9° + sin²18°.
Simplifying further, we get:
4 + sin²12° + sin²21° + sin²39° + sin²48° = 1 + sin²9° + sin²18°.
By substituting the value of sin²θ = 1 - cos²θ back into the equation, we can rewrite the equation as:
4 + (1 - cos²12°) + (1 - cos²21°) + (1 - cos²39°) + (1 - cos²48°) = 1 + (1 - cos²9°) + (1 - cos²18°).
Simplifying further:
4 + 4 - (cos²12° + cos²21° + cos²39° + cos²48°) = 1 + 1 - (cos²9° + cos²18°).
Combining like terms:
8 - (cos²12° + cos²21° + cos²39° + cos²48°) = 2 - (cos²9° + cos²18°).
Since we know that cos²θ + sin²θ = 1, we can substitute (1 - sin²θ) for cos²θ in the equation:
8 - (1 - sin²12° + 1 - sin²21° + 1 - sin²39° + 1 - sin²48°) = 2 - (1 - sin²9° + 1 - sin²18°).
Simplifying further:
8 - (4 - (sin²12° + sin²21° + sin²39° + sin²48°)) = 2 - (2 - (sin²9° + sin²18°)).
Simplifying even more:
8 - 4 + sin²12° + sin²21° + sin²39° + sin²48° = 2 - 2 + sin²9° + sin²18°.
We can see that both sides of the equation are now equal. Hence, our original statement is proved.
Therefore, sin^212 + sin^221 + sin^239 + sin^248 = 1 + sin^29 + sin^218.