Answer:
The solutions are 0° and 3π
Step-by-step explanation:
On solving the equation given;
<u>Since sin is negative in the 3rd and 4th quadrant, </u>
In the 3rd quadrant;
x =180°+2π
x = π + 2π
x = 3π
In the 4th quadrant;
x = 360°-2π
x = 2π-2π
x = 0°
I hope this helped, have a good day
Y= x-5 is equivelent to the equation
Answer:1630.72
Step-by-step explanation:V=bhl 11.2*11.2*13=1630.72
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
