WEIGHTAGE FOR CLASS ---XI One Paper Three Hours Max Marks. 100 Units Mark s I. SETS AND FUNCTIONS II. ALGEBRA III. COORDINATE GEOMETRY IV. CALCULUS V. MATHEMATICAL REASONING VI. STATISTICS AND PROBABILITY 29 37 13 06 03 12 100 Fundamental Trigonometric Identities
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WEIGHTAGE FOR CLASS ---XI
One Paper Three Hours Max Marks. 100
Units MarksI. SETS AND FUNCTIONS
II. ALGEBRA
III. COORDINATE GEOMETRY
IV. CALCULUS
V. MATHEMATICAL REASONING
VI. STATISTICS AND PROBABILITY
29
37
13
06
03
12 100
Fundamental Trigonometric Identities
Before we start to prove trigonometric identities, we see where the basic identities come from.
Recall the definitions of the reciprocal trigonometric functions, csc θ, sec θ and cot θ
from the trigonometric functions chapter:
After we revise the fundamental identities, we learn about: Proving trigonometric identities
Now, consider the following diagram where the point (x, y) defines an angle θ at the origin,
and the distance from the origin to the point is r units:
From the diagram, we can see that the ratios sin θ and cos θ are defined as:
and
Now, we use these results to find an important definition for tan θ:
Now, also so we can conclude that:
Also, for the values in the diagram, we can use Pythagoras' Theorem and obtain:
y2 + x2 = r2
Dividing through by r2 gives us:
so we obtain the important result:
sin2 θ + cos2 θ = 1
We now proceed to derive two other related formulas that can be used when proving trigonometric identities.
It is suggested that you remember how to find the identities, rather than try to memorise each one.
Dividing sin2 θ + cos2 θ = 1 through by cos2 θ gives us:
so
tan2 θ + 1 = sec2 θ
Dividing sin2 θ + cos2 θ = 1 through by sin2 θ gives us:
so
1 + cot2 θ = csc2 θ
Trigonometric Identities Summary
Proving Trigonometric Identities
Suggestions...
1. Learn well the formulas given above (or at least, know how to find them quickly).
The better you know the basic identities, the easier it will be to recognise what is going on in the problems.
2. Work on the most complex side and simplify it so that it has the same form as the simplest side.
Don't assume the identity to prove the identity.
This means don't work on both sides of the equals side and try to meet in the middle.
3. Start on one side and make it look like the other side.
4. Many of these come out quite easily if you express everything on the most complex side in terms of sine and cosine only.
5. In most examples where you see power 2 (that is, 2 ), it will involve using the identity sin2 θ + cos2 θ = 1 (or one of the other 2 formulas that we derived above).
Using these suggestions, you can simplify and prove expressions involving trigonometric identities.
Prove that
sin y + sin y cot2 y = cosec y
Answer
FunctionAbbreviation
Description
Identities (using radians)
Sine sin
Cosine cos
Tangent tan (or tg)
Cotangent
cot (or ctg or ctn)
Secant sec
Cosecantcsc (or cosec)
Note that these values can easily be memorized in the form
If Q(x,y) is the point on the circle where the string ends,
we may think of as being an angle by associating to it the central angle with vertex O(0,0) and sides passing through the points P and Q. If instead of wrapping a length s of string around the unit circle, we decide to wrap it around a circle of radius R, the angle (in radians) generated in the process will satisfy the following relation:
Observe that the length s of string gives the measure of the angle only when R=1.
As a matter of common practice and convenience, it is useful to measure angles in degrees, which are defined by partitioning one whole revolution into 360 equal parts, each of which is then called one degree. In this way, one whole revolution around the unit circle measures radians and also 360 degrees (or ), that is:
Each degree may be further subdivided into 60 parts, called minutes, and in turn each minute may be subdivided into another 60 parts, called seconds:
Angle sum identities
Sine
Illustration of the sum formula.
Draw the angles α and β. Place P on the line defined by α + β at unit distance from the origin.
This will only be equal to the previous expression we got, if the imaginary and real parts are equal respectively. Hence we get:
cos(α + β) = cosαcosβ − sinαsinβ
sin(α + β) = sinαcosβ + sinβcosα ]
Tangent and cotangent
From the sine and cosine formulae, we get
Dividing both numerator and denominator by cos α cos β, we get
Similarly (using a division by sin α sin β), we get
Double-angle identities
From the angle sum identities, we get
and
The Pythagorean identities give the two alternative forms for the latter of these:
The angle sum identities also give
**It can also be proved using Eulers Formula
Mulitplying the exponent by two yields
But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields
It follows that
By multiplying we get
Because the imaginary and real parts have to be the same, we are left with the original identities
Half-angle identities
The two identities giving alternative forms for cos 2θ give these:
One must choose the sign of the square root properly—note that if 2π is added to θ the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore the correct sign to use depends on the value of θ.
tan function, we have
If we multiply the numerator and denominator inside the square root by (1 + cos θ), and do a little manipulation using the Pythagorean identities, we get
If instead we multiply the numerator and denominator by (1 - cos θ), we get
This also gives
Similar manipulations for the cot function give
Example. verify the identity
Answer. We have
which gives
But
and since
and , we get finally
Remark. In general it is good to check whether the given formula is correct. One way to do that is to substitute some numbers for the variables. For example, if we take a=b = 0, we get
or we may take . In this case we have
Example. Find the exact value of
Answer. We have
Hence, using the additions formulas for the cosine function we get
Since
we get
Example. Find the exact value for
Answer. We have
Since
we get
Finally we have
Remark. Using the addition formulas, we generate the following identities
Double-Angle and Half-Angle formulas are very useful. For example, rational functions of sine and cosine wil be very hard to integrate without these formulas. They are as follow
Example. Write as an expression involving the trigonometric functions with their first power.
Answer. We have
Hence
Since , we get
or
Example: Verify the identity
Answer. We have
Using the Double-Angle formulas we get
Putting stuff together we get
From the Double-Angle formulas, one may generate easily the Half-Angle formulas
In particular, we have
Example. Use the Half-Angle formulas to find
Answer. Set . Then
Using the above formulas, we get
Since , then is a positive number. Therefore, we have
Same arguments lead to
Example. Check the identities
Answer. First note that
which falls from the identity . So we need to verify only one identity. For example, let us verify that
using the Half-Angle formulas, we get
which reduces to
Product and Sum Formulas
From the Addition Formulas, we derive the following trigonometric formulas (or identities)
Remark. It is clear that the third formula and the fourth
are identical (use the property to see it).
The above formulas are important whenever need rises to transform the product of sine and cosine into a sum. This is a very useful idea in techniques of integration.
Example. Express the product as a sum of trigonometric functions.
Answer. We have
which gives
Note that the above formulas may be used to transform a sum into a product via the identities
= 2cos 4 A (sin 5 A−sin 9 A)2 sin 4 A(sin 5 A−sin 9 A)
= cot 4A.
Question – 10 Prove that √2+√2+√2+2 cos 8θ = 2COSθ. Answer: √2+√2+√2(1+cos8 θ) = √2+√2+√2×2cos ²4 θ
√2+√2(1+cos 4 θ) = √2+√2× 2cos ²2θ = √2(1+cos2θ) = 2cosθ. [By using 1+cos2 θ= 2cos2θ] Question – 11 Find the general solution of the following equation:
4sinxcosx+2sinx+2cosx+1 = 0
Answer: Above equation can be written as (4sinxcosx+2sinx) + (2cosx+1 ) = 0
⇨ 2sinx (2cosx+1) + (2cosx+1) = 0
⇨ (2sinx+1) (2cosx+1) = 0
⇨ (2sinx+1) = 0 or 2cosx+1 = 0
⇨ sinx = -1/2 or cosx = -1/2
⇨ sinx = sin(π + π6 )
or cosx = cos (π - π3 )
⇨ x = nπ +(-1)n 7 π6
or x = 2nπ ± 2 π3 , ∀ n Є Z (Integers).
Question – 12 If cosx = -1213 and π <x <3 π
2 , find the
value of sin3x and cos3x.
Answer: Since x lies in 3rd quadrant ∴ sinx is negative.
Sinx = – √1−cos ² x = - 513 then sin3x = 3sinx
– sin3x = -20352197
Cos3x = 4cos3x – 3cosx = - 8282197 [by
putting the values of sinx & cosx]
Question – 13 (i) If cos(A+B) sin(C-D) = cos(A-B) sin(C+D) , then show that
tanA tanB tanC + tanD = 0
(ii) If sinθ = n sin(θ+2𝛂), prove that
tan(θ+𝛂) = 1+n1−n tan𝛂.
Answer: We can write above given result as cos (A+B)cos( A−B) = sin(C+D)
sin(C−D)
By C & D cos ( A+B )+cos ( A−B)cos ( A+B )−cos( A−B) =
(c) Find in degrees and radians the angle subtended b/w the hour hand and the minute hand
Of a clock at half past three. [answer is 750 , 5π/12]
(d) 1tan 3 A−tanA - 1
cot 3 A−cotA = cot 2A. [Hint
put cot3A = 1/tan3A & cotA = 1/tanA then take l.c.m.
and use formula tan(A-B).]
(e) If α , 𝛃 are the distinct roots of acosθ +
bsinθ = c, prove that sin(𝛂+𝛃) = 2 aba ²+b ² .
Solution of (e) If α , 𝛃 are the distinct roots of acosθ + bsinθ = c, then
acos𝛂 + bsin𝛂 = c & acos𝛃 + bsin𝛃 = c
By subtracting , we get a(cos𝛂 – cos𝛃) + b(sin𝛂 – sin𝛃) = 0 ⇨ a(cos𝛂 – cos𝛃) = b(sin𝛂 – sin𝛃) ⇨ 2a sin α+β2 sin α−β
2 = 2b cosα+β2 sin
α−β2
⇨ tanα+β2 = b
a ⇨ sin(𝛂+𝛃) = 2 tan
α+β2
1+ tan ²α+β
2
⇨
sin(𝛂+𝛃) = 2aba ²+b ² .
Question: If sin2A = k sin2B, prove that tan( A+B)tan (A−B) =
k+1k−1
[ Hint by c & d sin 2 A+sin 2 B
sin 2 A−sin 2 B = k+1k−1 , use formula of sinx+siny= 2sin( x+ y
2¿ cos¿)]
Question: If cos(x+2A) = n cosx, show that cotA = n+11−n tan(x+A). Question: Prove that tan(∝−β ¿ = sin 2 β
5−cos 2 β , if 2tan∝ = 3tanβ. [Hint: L.H.S. = tanα−tanβ1+ tanα . tanβ , put the value of tan∝ and simplify it]. Question: If sinx + siny = a and cosx + cosy = b,find the
value of cos(x-y).
Answer: squaring and adding above results, we will get cos(x-y) = ½[a2+b2 – 2]
Question: Show that 1sin 10° - √ 3
cos10 ° =4
[Hint: 2 [ 1/2sin 10° - √3/2
cos10 ° ] = 2 [ sin 30°sin 10° - cos30 °
cos10 ° ]
Question: Prove that : 1+sinx−cosx1+sinx+cosx = tan(x/2)
[ Hint: use 1 – cosx = 2sin2x/2 , 1+cosx = 2cos2x/2 and sinx = 2sinx/2.cosx/2]