# MATHEMATICAL TABLES: CONTAINING THE COMMON, HYPERBOLIC, AND LOGISTIC

The sin of angle difference identity is a trigonometric identity. It’s used khổng lồ expand sin of subtraction of two angles functions such as $sin(A-B)$, $sin(x-y)$, $sin(alpha-eta)$, và so on. You know how to expand sin of difference of two angles và it’s essential to lớn learn how it is derived in mathematical khung in trigonometry.

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## Proof

The sine of difference of two angles formula is derived in trigonometry by the geometrical approach. Geometrically, a right triangle is constructed and it contains difference of two angles. Then, the angle difference identity is derived in mathematical form on the basis of this triangle.

### Construction of triangle with difference of two angles

$Delta EDF$ is a right triangle & the $angle EDF$ is taken as $x$.

$(1). ,,,$ Draw a line from point $D$ khổng lồ side $overlineEF$. It divides the angle $x$ as two angles và intersects the side $overlineEF$ at point $G$. If $angle GDE$ is $y$, then $angle GDF = x-y$ geometrically.

$(2). ,,,$ Take set square and draw a perpendicular line khổng lồ side $overlineDE$ but it should intersect the point $G$. Take, the perpendicular intersection point as $H$. In this case, $overlineDE perp overlineGH$.

$(3). ,,,$ Draw a perpendicular line to side $overlineDF$ from $H$ & it intersects the side $overlineDG$ at point $I$ và also perpendicularly intersects the side $overlineDF$ at point $J$.

$(4). ,,,$ Draw a perpendicular line to side $overlineHJ$ from point $G$ & it intersects the side $overlineHJ$ at point $K$.

The geometrical steps formed three right triangles, and they’re $Delta GDH$ with angle $y$, $Delta FDG$ with angle $x-y$ và $Delta KHG$ with an unknown angle.

### Express Sine of difference of angles in its ratio form

In $Delta GDF$, the $angle GDF = x-y$. The sine of difference of two angles is written as $sin(x-y)$ và it can be written in its ratio khung on the basis of this triangle.

$sin(x-y) ,=, dfracFGDG$

The sides $overlineFG$ and $overlineKJ$ are parallel lines & they’re equal. Therefore, $FG = KG$.

$implies$ $sin(x-y) ,=, dfracKJDG$

The perpendicular side $overlineKG$ divides the side $overlineHJ$ as two sides $overlineHK$ và $overlineKJ$.

Therefore $HJ = HK+KJ$ $implies KJ = HJ-HK$

The length of the side $overlineKJ$ can be replaced by its equivalent value in the expansion of sin of angle difference function.

$implies sin(x-y) ,=, dfracHJ-HKDG$

$implies sin(x-y) ,=, dfracHJDG-dfracHKDG$

### Convert Lengths of sides as trigonometric functions

$HJ$ is perpendicular of $Delta HDJ$ & the $angle HDJ = x$. Try khổng lồ express $HJ$ in a trigonometric function as per this triangle.

$sinx ,=, dfracHJDH$ $implies$ $HJ ,=, DHsinx$

Now, replace the $HJ$ by its equivalent value in the expansion of $sin(x-y)$ function.

$implies sin(x-y)$ $,=,$ $dfracDHsinxDG$ $-$ $dfracHKDG$

$implies sin(x-y)$ $,=,$ $dfracDHDG imes sinx$ $-$ $dfracHKDG$

$DH$ và $DG$ are lengths of adjacent side & hypotenuse in the $Delta HDG$. The $angle HDG$ is $y$. Now, express relationship between them in the size of a trigonometric function.

$cosy ,=, dfracDHDG$

Therefore, the ratio of $DH$ to lớn $DG$ can be replaced by $cosy$ in the expansion of sin of subtraction of two angles function.

$implies sin(x-y)$ $,=,$ $cosy imes sinx$ $-$ $dfracHKDG$

$implies sin(x-y)$ $,=,$ $cosysinx$ $-$ $dfracHKDG$

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $dfracHKDG$

### Find unknown angle of right triangle

$HK$ is length of the adjacent side in the $Delta KHG$ but its angle is unknown and it’s not possible to lớn express $HK$ in the size of a trigonometric function. So, let’s find the $angle KHG$ geometrically to lớn continue the derivation of sin of angle difference identity.

Two angles are known but an angle is unknown in the $Delta JDI$. The $angle JDI = x-y$, $angle DJI = 90^circ$ và $angle DIJ$ is unknown. It can be evaluated by using sum of angles rule of a triangle.

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$angle JDI$ $+$ $angle DIJ$ $+$ $angle IJD$ $,=,$ $180^circ$

$implies$ $x-y$ $+$ $angle DIJ$ $+$ $90^circ$ $,=,$ $180^circ$

$implies$ $angle DIJ$ $,=,$ $180^circ-90^circ-x+y$

$,,, herefore ,,,,,,$ $angle DIJ$ $,=,$ $90^circ-x+y$

The $angle DIJ$ và $angle GIH$ are vertically opposite angles. Therefore, they are equal geometrically.

$angle GIH$ $,=,$ $angle DIJ$

$,,, herefore ,,,,,, angle GIH$ $,=,$ $90^circ-x+y$

The $angle DIG$ is a straight angle but it’s divided as two angles $angle DIH$ & $angle GIH$ by the intersection of the side $overlineIH$.

Therefore, $angle DIG ,=, 180^circ$

$implies$ $angle DIH$ $+$ $angle GIH$ $,=,$ $180^circ$

It is already derived that $angle GIH$ $,=,$ $90^circ-x+y$

$implies$ $angle DIH$ $+$ $90^circ-x+y$ $,=,$ $180^circ$

$implies$ $angle DIH$ $,=,$ $180^circ-90^circ+x-y$

$,,, herefore ,,,,,,$ $angle DIH$ $,=,$ $90^circ+x-y$

Therefore, it’s derived that $angle DIH$ is equal khổng lồ $90^circ+x-y$ but it’s one of the angles of the $Delta IDH$.

In $Delta IDH$, the $angle HDI = y$, $angle DIH$ $,=,$ $90^circ+x-y$ but $angle IHD$ is unknown. It can be derived by using sum of angles of a triangle formula.

$angle DIH$ $+$ $angle IHD$ $+$ $angle HDI$ $,=,$ $180^circ$

$implies$ $90^circ+x-y$ $+$ $angle IHD$ $+$ $y$ $,=,$ $180^circ$

$implies$ $angle IHD$ $+$ $90^circ+x-y+y$ $,=,$ $180^circ$

$implies$ $angle IHD$ $+$ $ equirecancel 90^circ+x-cancely+cancely$ $,=,$ $180^circ$

$implies$ $angle IHD$ $,=,$ $180^circ-90^circ-x$

$,,, herefore ,,,,,,$ $angle IHD$ $,=,$ $90^circ-x$

Actually, the sides $overlineDH$ & $overlineHG$ are perpendicular lines. So, the angle between them is a right angle.

Therefore, $angle GHD$ $,=,$ $90^circ$

The side $overlineIH$ divides the right angle ($angle GHD$) as two angles $angle IHD$ và $angle IHG$ but the $angle IHD$ is derived as $90^circ-x$.

$implies$ $angle IHD$ $+$ $angle IHG$ $,=,$ $90^circ$

$implies$ $90^circ-x$ $+$ $angle IHG$ $,=,$ $90^circ$

$implies$ $angle IHG$ $,=,$ $90^circ-90^circ+x$

$implies$ $angle IHG$ $,=,$ $ equirecancel cancel90^circ-cancel90^circ+x$

$,,, herefore ,,,,,,$ $angle IHG$ $,=,$ $x$

The $angle IHG$ & $angle KHG$ are congruent angles.

$,,, herefore ,,,,,,$ $angle KHG = x$

Therefore, it is calculated that the angle of $Delta KHG$ is $x$. Actually, the angle of $Delta EDF$ is also $x$. It is possible because the $Delta EDF$ and $Delta KHG$ are similar triangles.

### Convert Lengths of sides as trigonometric functions

The angle of $Delta KHG$ is known and the proof of $sin(x-y)$ can be continued further.

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $dfracHKDG$

The $HK$ can be expressed in the khung a trigonometric function.

$cosx ,=, dfracHKHG$

$implies$ $HK ,=, HGcosx$

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $dfracHGcosxDG$

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $dfracHGDG imes cosx$

$HG$ and $DG$ are lengths of the opposite side & hypotenuse in the $Delta HDG$. The angle of this triangle is $y$. The relationship of angle with the ratio between them can expressed in the form of a trigonometric function.

$siny ,=, dfracHGDG$

Now, substitute the ratio of $HG$ to $DG$ by its equivalent value in the expansion of sine of difference of two angles.

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $siny imes cosx$

$implies sin(x-y)$ $,=,$ $sinxcosy$ $-$ $sinycosx$

$,,, herefore ,,,,,,$ $sin(x-y)$ $,=,$ $sinxcosy$ $-$ $cosxsiny$

Therefore, it is proved that sine of difference of two angles is equal khổng lồ subtraction of products of sin và cosine of the two angles.

It is called sin of angle difference identity or sin of subtraction of two angles và used as a formula in mathematics.

Other formsThe sin of angle difference identity can also be written in several ways.

For example, if $A$ và $B$ are angles, then the sin of subtraction of two angles is $sin(A-B)$ but it is expanded in the same form.

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$(1) ,,,$ $sin(A-B)$ $,=,$ $sinAcosB$ $-$ $cosAsinB$

Similarly, if $alpha$ and $eta$ are two angles, then the sine of difference of two angles is written as $sin(alpha-eta)$ & it is expanded as follows.

$(2) ,,,$ $sin(alpha-eta)$ $,=,$ $sinalphacoseta$ $-$ $cosalphasineta$