Trigonometry Help: Solving for Theta

[ibysaiyan]

1.
Hi
Its the bit (c) where i am stuck at although it doesn't look much complicated, for all i know is that max value for cos =1 , so cos inverse becomes 0. The answer on the mark scheme is theta = 326 which i can't figure out.Thanks

Homework Equations

The Attempt at a Solution

 

[kg4pae]

1.
Hi
Its the bit (c) where i am stuck at although it doesn't look much complicated, for all i know is that max value for cos =1 , so cos inverse becomes 0. The answer on the mark scheme is theta = 326 which i can't figure out.Thanks

Homework Equations

The Attempt at a Solution

I'm just wondering, have you tried it? I would start by observing that

cos(\theta + \alpha) = cos(\theta) cos(\alpha) - sin(\theta) sin(\alpha)

After solving for \alpha, you then can find a maximum cos(\theta+\alpha) which when multiplied by R will give the answer to part (c).

If you need further help, post your work so far.

 

[ibysaiyan]

I'm just wondering, have you tried it? I would start by observing that

cos(\theta + \alpha) = cos(\theta) cos(\alpha) - sin(\theta) sin(\alpha)

After solving for \alpha, you then can find a maximum cos(\theta+\alpha) which when multiplied by R will give the answer to part (c).

If you need further help, post your work so far.

Ah how weird i just did another similar question of which i got the answers hmm anyway, here is what i did:
i got the R value square root. 13 (by equating co-efficients of sin and cos),
theta 33.7.
P.S: Sorry i have yet to become latex friendly , if only someone could post me a tutorial on how to use it lol .

 

[kg4pae]

I'm just wondering, have you tried it? I would start by observing that

cos(\theta + \alpha) = cos(\theta) cos(\alpha) - sin(\theta) sin(\alpha)

After solving for \alpha, you then can find a maximum cos(\theta+\alpha) which when multiplied by R will give the answer to part (c).

If you need further help, post your work so far.

Note that cos(\theta+\alpha)=1 means that \theta=-\alpha. However, the pattern repeats every 360^o.

 

[ibysaiyan]

I'm just wondering, have you tried it? I would start by observing that

cos(\theta + \alpha) = cos(\theta) cos(\alpha) - sin(\theta) sin(\alpha)

After solving for \alpha, you then can find a maximum cos(\theta+\alpha) which when multiplied by R will give the answer to part (c).

If you need further help, post your work so far.

Oh! would i let Rcos(\theta+\alpha)= 1?

 

[kg4pae]

Ah how weird i just did another similar question of which i got the answers hmm anyway, here is what i did:
i got the R value square root. 13 (by equating co-efficients of sin and cos),
theta 33.7.
P.S: Sorry i have yet to become latex friendly , if only someone could post me a tutorial on how to use it lol .

No problem. A good start with LaTex is http://frodo.elon.edu/tutorial/tutorial/". Others can be found by Googling "latex tutorial". At any rate, take care, 73s and clear skies.

 

[ibysaiyan]

No problem. A good start with LaTex is http://frodo.elon.edu/tutorial/tutorial/". Others can be found by Googling "latex tutorial". At any rate, take care, 73s and clear skies.

Thanks a lot!:) for the link .

 

[kg4pae]

Oh! would i let Rcos(\theta+\alpha)= 1?

Not quite. Let cos(\theta+\alpha)=1. That will give the maximum for 3 cos(\theta) - 2 sin(\theta) after it is multiplied by R. Since R is effectively a constant any maximum of cos(\theta+\alpha) will be proportionate to R cos(\theta+\alpha)=3 cos(\theta) - 2 sin(\theta).