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Sin 90 + theta

  1. Jul 8, 2012 #1
    i understand how sin90-theta is cos , but i am having trouble with sin 90+theta = cos ...please explain
  2. jcsd
  3. Jul 8, 2012 #2


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    I'm not clear what you are asking. The left side of each equation, sin(90- theta), and sin(90+ theta) make sense but the right side "cos" is meaningless. Do you mean cos(theta)?

    What definition of sine and cosine are you using? If you are using the basic trigonometry definitions, then if one angle in a triangle is theta, the other is 90- theta so that "near side" and "opposite side" are reversed so that "sine" an "cosine" are switched. But in that case "sin(90+ theta)" makes no sense. A right triangle cannot have an angle larger than 90 degrees.

    In Calculus, we need sine and cosine functions for all x and so use more general definitions. One common one is this:

    Draw the unit circle on a coordinate system ("the unit circle" has center at (0, 0) and radius 1). Given positive number t, measure around the circumference of the circle counter-clockwise (for negative t, clockwise). The coordinates of the terminal point are, by definition, (cos(t), sin(t)). "By definition" here means the definition of sine and cosine- whatever those coordinates are, we define cos(x) and sin(x) to be those values)

    Of course, to get "sin(90)= 1" and "cos(90)= 0" we have to interpret "t" in radians. The unit circle has circumference [itex]2\pi[/itex] so that 1/4 of the way around, from (1, 0) to (0, 1) is [itex]2\pi/4= \pi/2[/itex] radians which is equivalent to 90 degrees.

    Now, notice that the points on the unit circle corresponding to central angles 90- theta and 90+ theta (distances along the circumference [itex]\pi/2- x[/itex] and [itex]\pi/2+ x[/itex] with x in radians) are symmetric about the y-axis. They have the same y coordinate, and their x coordinates differ only in sign. Thus [itex]sin(x+ \pi//2)= sin(x- \pi/2)[/itex] where x is in radians or [itex]sin(90+ \theta)= sin(90- \theta)[/itex] with [itex]\theta[/itex] in degrees.
  4. Jul 8, 2012 #3
    yes i meant cos theta

    im using the basic trig identities .... in my textbook we have a table which denotes ....

    sin(90-theta) = cos heta
    sin(90+theta) = cos theta
    sin(180-theta) = sin theta
    sin(180+theta) = - sin
    and so on..
    i was also confused because trig is applicable only in right angles ... i cant make sence of any except the first .
  5. Jul 8, 2012 #4
    Hello Sam,

    Firstly to echo what Halls of Ivy said,
    The angle theta can be as large as we like.
    I'm sure you know that there are 360 degrees in a whole circle 720 in two circles etc.
    For every whole circle we wind round we take off 360 degrees.

    So 750 degrees is equivlent to 750 -360-360 =30 deggrees

    By equivalent to I mean that if you wind a clock hand round 750 degrees or 30 degrees it will end up in the same place.

    So it is reasonable to think that the values of trig functions for 750 will be the same as those for 30.
    This is indeed true.

    The only other thing to understand is that we must agree which way to wind the clock handle and where from.

    In mathematics we start from the positive x axis and wind anticlockwise.
    That is the convention.

    To look more closely at this look now at my sketches.

    The first shows the clock hand winding round from the x axis an angle theta.
    This hand also creates two triangles with the x axes and I have written out the sin and cos for both triangles.
    This shows the relationship between theta and (90 - theta)

    The second sketch shows what happens with 180 - theta.
    Can you form the triangles for yourself here?
    Remember that y is still positive, but x is now negative.

    The second attachment shows what happens when you use all four quadrants.
    Some of the trig quantities are negative and some positive in any quadrant, but there is always one positive in any quadrant.
    I have worked out the tangent in this instance.

    Can you do the sin and cos now?

    Attached Files:

  6. Jul 8, 2012 #5
    hey Studiot , thank you for your reply , it solved most of my difficulties but i still have a confusion

    in first quadrant , we get cos theta = x/underoot x^2 + y^2

    but then if we do it for 90+theta .. my value for sin (90 + theta) is -x/underoot x^2 + y^2

    which is basically our previous value of cos theta * (-)

    therefore sin 90 + theta should be equal to - cos theta , not cos theta
  7. Jul 8, 2012 #6


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    Yes, that is true. "[itex]cos(\theta)[/itex]" is the x-component of the [itex](cos(\theta), sin(\theta))[/itex] and so is negative as we move from first quadrant to second quadrant.

    However [itex]sin(\theta)[/itex] is the y component and does NOT change sign. That is, if [itex]0<\theta< 90[/itex], in the first quadrant, [itex]\theta+ 90[/itex] is in the second quadrant. [itex]sin(\theta+ 90)[/itex] is still positive as is [itex]cos(\theta)[/itex].
  8. Jul 8, 2012 #7
    Hello again Sam.

    Supplementary angles add to 180

    Complementary angles add to 90

    Complementary angles are more tricky since you have to exchange x and y or sin and cos in the formulae to get the correct triangle.
    Sin is still OHMS and cos is still AHMS however.
    You do not do this with supplementary angles.

    To see why look at the attachment.

    The first quadrant triangle with angle theta is shown as OAB.

    I have generated a line OD in the second quadrant by adding 90 to theta.

    However the triangle you want is not ODC.

    The triangle you want is ODE as shown.

    Using this triangle the attachment shows that

    cos(90 + θ) = -sin(θ)

    You should always use supplementary angles where possible for this reason.

    go well

    Attached Files:

  9. Jul 8, 2012 #8
    ty studiot , i understod itt now , btw if u dont mind me asking , how was my logic wrong ? the one which proved sin 90+theta = - cos theta
  10. Jul 8, 2012 #9
    If you don't mind, can you post the full table? I would like to memorize these but can't find these on the internet. Thx
  11. Jul 8, 2012 #10
  12. Jul 9, 2012 #11

    You used the wrong triangle. That's why I said complementary angles are more tricky.

    It's in the last attachment x/√(x2+y2) is for triangle ODC
  13. Jul 9, 2012 #12
    I do not see why we are having such diffuculty.
    From the angle sum formulae of the sine function, we obtain
    [tex]\sin(\pi/2 + \theta) = \sin(\pi/2)\cos(\theta) + \cos(\pi/2)\sin(\theta) = \cos(\theta)[/tex]
    since [itex]\sin(\pi/2)=1[/itex] and [itex]\cos(\pi/2)=0[/itex].
  14. Jul 9, 2012 #13
    i wanted to prove it logically
  15. Jul 9, 2012 #14
    Sam I owe you an apology, there was an error in my post7, so well spotted.

    I tried this morning but couldn't amend it however I will supply a correction later tonight.
  16. Jul 9, 2012 #15
    OK I promised an ammendment.

    You notices that I had written sin(90+θ) instead of sin(90-θ).

    In fact it makes no difference they are the same.

    However since this was an unintentional slip and I should have used sin(90-θ) here is a geometric proof that

    cos(90+θ) = -sin(θ)

    in attachment 1.

    Since is is some effort I will leave you to construct similar proofs for sin and tan.

    It is really much easier to use Millenial's method for this and I have shown all three in attachment 2

    go well

    Attached Files:

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