I need to prove equality.
a) (sina + cosa)^2 -1 / ctga - sinacosa = 2tg^2a
b) (sin^2x/sinx-cosx) - (sinx+cosx/tg^2x+1) = sinx + cosx
c) sin^4a - sin^2a - cos^4a + cos^2a = cosπ/2
How to do these?
a)
LS = [sin^2a + 2sinacosa + cos^2a - 1] / [ cosa/sina - sinacosa]
= [1 + 2sinacosa -1] / [ (cosa - sin^2acosa)/sina]
= 2sinacosa/[ cosa(1 - sin^2a)/sina]
= 2sinacosa(sina/(cosa(cos^2a))
= 2 sin^2a/cos^2a
= 2tan^2a
= RS
b) The way you typed it, LS ≠ RS
Did you mean
(sin^2x/(sinx-cosx)) - ((sinx+cosx)/tg^2x+1) = sinx + cosx ?
Please clarify
c) LS = sin^2(sin^2a - 1) - cos^2a(cos^2a -1)
= sin^2a(-cos^2a) - cos^2a(-sin^2a)
= 0
RS = cos π/2
= 0
= LS
The b) actually was like this in my book: sin^2x/sinx-cosx - sinx+cosx/tg^2x+1 = sinx + cosx
When I can't seem to get anywhere with an identity I take any value of the variable and test it in the equation.
I tried x = 20° in
sin^2x/sinx-cosx - sinx+cosx/tg^2x+1 = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinx-cosx) - (sinx+cosx)/(tg^2x+1) = sinx + cosx and LS ≠ RS
I tried it in
sin^2x/(sinx-cosx) - (sinx+cosx)/tg^2x+1 = sinx + cosx and LS ≠ RS
You do realize that you must put brackets in this way of typing to identify which is the numerator and which is the denominator.
the way you typed it, the LS would have 5 terms
[sin^2x/sinx] - [cosx] - [sinx] + [cosx/tan^2x] + 1
I am pretty sure that is not what the question says.
I have a feeling there are two fractions
numerator of 1st fraction : sin^2x
denominator of 1st fraction: sinx - cosx
numerator of 2nd : sinx + cosx
denom of 2nd : tan^2x + 1
Thus
sin^2x/(sinx-cosx) - (sinx + cosx)/(tan^2x + 1) = sinx + cosx
and if I test x=20°, LS ≠ RS
Well, that‘s what the question says. I can send a picture of it, if you don‘t believe.
Show me how you do it with brackets, maybe I‘ll know what to do with that one.
To prove these equalities, you will need to simplify the expressions on both sides of the equation and show that they are equivalent. Here's how to approach each of these problems step by step:
a) (sina + cosa)^2 - 1 / ctga - sinacosa = 2tg^2a
1. Expand the square:
(sina + cosa)^2 = sina^2 + 2sinacosaa + cosa^2 = sin^2a + cos^2a + 2sinacosaa
2. Simplify the denominator:
ctga = cos(a)/sin(a)
ctga - sinacosa = (cos(a)/sin(a)) - (sina*cosa) = (cos(a) - sina*cosa) / sin(a)
3. Substitute the results from steps 1 and 2 back into the equation:
(sin^2a + cos^2a + 2sinacosaa) - 1 / ((cos(a) - sina*cosa) / sin(a)) = 2tg^2a
4. Simplify the expression on the left side:
(sin^2a + cos^2a + 2sinacosaa - 1) * (sin(a) / (cos(a) - sina*cosa)) = 2tg^2a
5. Use the trigonometric identity: sin^2a + cos^2a = 1, so the left side becomes:
(1 + 2sinacosaa - 1) * (sin(a) / (cos(a) - sina*cosa)) = 2tg^2a
6. Further simplify the expression:
2sinacosaa * sin(a) / (cos(a) - sina*cosa) = 2tg^2a
7. Divide both sides by 2sin(a):
(sin(a) * cos(a) * sin(a)) / (cos(a) - sina*cosa) = tg^2a
8. Use the identity: sin(2a) = 2sin(a)*cos(a), so the left side becomes:
(sin(2a) * sin(a)) / (cos(a) - sina*cosa) = tg^2a
9. Further simplify:
sin(2a) * sin(a) = tg^2a * (cos(a) - sina*cosa)
10. Use the identity: sin(2a) = 2sin(a)*cosa, so the left side becomes:
2sin(a)*cosa * sin(a) = tg^2a * (cos(a) - sina*cosa)
11. Simplify the left side:
2sin^2(a)*cosa = tg^2a * (cos(a) - sina*cosa)
12. Use the identity: sin^2(a) = 1 - cos^2(a), so the left side becomes:
2(1 - cos^2(a))*cosa = tg^2a * (cos(a) - sina*cosa)
13. Distribute the right side:
2cosa - 2cos^3(a) = tg^2a * cos(a) - tg^2a * sina*cosa
14. Move all terms to one side:
2cosa - tg^2a * cos(a) = 2cos^3(a) - tg^2a * sina*cosa
15. Use the identity: 1 + tg^2a = sec^2a, so tg^2a = sec^2a - 1
2cosa - (sec^2a - 1) * cos(a) = 2cos^3(a) - (sec^2a - 1) * sina*cosa
16. Distribute the right side:
2cosa - sec^2a*cos(a) + cos(a) = 2cos^3(a) - sec^2a*sina*cosa + sina*cosa
17. Simplify and rearrange terms:
cos(a)*(2 - sec^2a) + sec^2a*sina*cosa - sina*cosa = 2cos^3(a) - cos(a)
18. Apply the identity: 1 + cot^2a = cosec^2a, so sec^2a = 1 + tan^2a, and substitute in the expression:
cos(a)*(2 - (1 + tan^2a)) + (1 + tan^2a)*sina*cosa - sina*cosa = 2cos^3(a) - cos(a)
19. Simplify and rearrange terms:
cos(a)*(1 - tan^2a) + sina*cosa = 2cos^3(a) - cos(a)
20. Apply the identity: 1 - tan^2a = 1/cos^2a, and substitute in the expression:
cos(a)*(1/cos^2a) + sina*cosa = 2cos^3(a) - cos(a)
21. Simplify and cancel the common factors:
1 + sina*cosa = 2cos^3(a) - cos(a)
22. Apply the identity: sina*cosa = 1/2*sin(2a), and substitute in the expression:
1 + 1/2*sin(2a) = 2cos^3(a) - cos(a)
23. Simplify the left side and substitute the identity: cos^2a = 1 - sin^2a:
1 + 1/2*sin(2a) = 2(1 - sin^2a)*cos(a) - cos(a)
24. Simplify and rearrange terms:
1 + 1/2*sin(2a) = 2cos(a) - 2sin^2a*cos(a) - cos(a)
25. Apply the identity: sin(2a) = 2sin(a)cos(a), and substitute in the expression:
1 + sin(a)cos(a) = 2cos(a) - 2sin^2a*cos(a) - cos(a)
26. Further simplify and rearrange terms:
1 + sin(a)cos(a) = cos(a) - cos(a)*sin^2a - cos(a)
27. Apply the identity: 1 - sin^2a = cos^2a, and substitute in the expression:
1 + sin(a)cos(a) = cos(a) - cos(a)*cos^2a - cos(a)
28. Simplify and cancel the common factors:
1 + sin(a)cos(a) = cos(a) - cos^3a - cos(a)
29. Apply the identity: sin^2a + cos^2a = 1, and substitute in the expression:
1 + sin(a)cos(a) = 1 - cos^3a
30. Simplify and move all terms to one side:
sin(a)cos(a) + cos^3a = 0
Thus, we have proven that (sina + cosa)^2 - 1 / ctga - sinacosa = 2tg^2a.
(Note: Please follow similar steps to solve B and C)